1106 lines
33 KiB
Ada
1106 lines
33 KiB
Ada
------------------------------------------------------------------------------
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-- --
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-- GNAT COMPILER COMPONENTS --
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-- --
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-- S Y S T E M . B I G N U M S --
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-- --
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-- B o d y --
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-- --
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-- Copyright (C) 2012-2015, Free Software Foundation, Inc. --
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-- --
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-- GNAT is free software; you can redistribute it and/or modify it under --
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-- terms of the GNU General Public License as published by the Free Soft- --
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-- ware Foundation; either version 3, or (at your option) any later ver- --
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-- sion. GNAT is distributed in the hope that it will be useful, but WITH- --
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-- OUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY --
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-- or FITNESS FOR A PARTICULAR PURPOSE. --
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-- --
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-- As a special exception under Section 7 of GPL version 3, you are granted --
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-- additional permissions described in the GCC Runtime Library Exception, --
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-- version 3.1, as published by the Free Software Foundation. --
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-- --
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-- You should have received a copy of the GNU General Public License and --
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-- a copy of the GCC Runtime Library Exception along with this program; --
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-- see the files COPYING3 and COPYING.RUNTIME respectively. If not, see --
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-- <http://www.gnu.org/licenses/>. --
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-- --
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-- GNAT was originally developed by the GNAT team at New York University. --
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-- Extensive contributions were provided by Ada Core Technologies Inc. --
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-- --
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------------------------------------------------------------------------------
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-- This package provides arbitrary precision signed integer arithmetic for
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-- use in computing intermediate values in expressions for the case where
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-- pragma Overflow_Check (Eliminate) is in effect.
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with System; use System;
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with System.Secondary_Stack; use System.Secondary_Stack;
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with System.Storage_Elements; use System.Storage_Elements;
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package body System.Bignums is
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use Interfaces;
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-- So that operations on Unsigned_32 are available
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type DD is mod Base ** 2;
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-- Double length digit used for intermediate computations
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function MSD (X : DD) return SD is (SD (X / Base));
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function LSD (X : DD) return SD is (SD (X mod Base));
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-- Most significant and least significant digit of double digit value
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function "&" (X, Y : SD) return DD is (DD (X) * Base + DD (Y));
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-- Compose double digit value from two single digit values
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subtype LLI is Long_Long_Integer;
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One_Data : constant Digit_Vector (1 .. 1) := (1 => 1);
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-- Constant one
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Zero_Data : constant Digit_Vector (1 .. 0) := (1 .. 0 => 0);
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-- Constant zero
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-----------------------
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-- Local Subprograms --
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-----------------------
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function Add
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(X, Y : Digit_Vector;
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X_Neg : Boolean;
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Y_Neg : Boolean) return Bignum
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with
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Pre => X'First = 1 and then Y'First = 1;
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-- This procedure adds two signed numbers returning the Sum, it is used
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-- for both addition and subtraction. The value computed is X + Y, with
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-- X_Neg and Y_Neg giving the signs of the operands.
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function Allocate_Bignum (Len : Length) return Bignum with
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Post => Allocate_Bignum'Result.Len = Len;
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-- Allocate Bignum value of indicated length on secondary stack. On return
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-- the Neg and D fields are left uninitialized.
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type Compare_Result is (LT, EQ, GT);
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-- Indicates result of comparison in following call
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function Compare
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(X, Y : Digit_Vector;
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X_Neg, Y_Neg : Boolean) return Compare_Result
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with
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Pre => X'First = 1 and then Y'First = 1;
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-- Compare (X with sign X_Neg) with (Y with sign Y_Neg), and return the
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-- result of the signed comparison.
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procedure Div_Rem
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(X, Y : Bignum;
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Quotient : out Bignum;
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Remainder : out Bignum;
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Discard_Quotient : Boolean := False;
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Discard_Remainder : Boolean := False);
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-- Returns the Quotient and Remainder from dividing abs (X) by abs (Y). The
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-- values of X and Y are not modified. If Discard_Quotient is True, then
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-- Quotient is undefined on return, and if Discard_Remainder is True, then
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-- Remainder is undefined on return. Service routine for Big_Div/Rem/Mod.
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procedure Free_Bignum (X : Bignum) is null;
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-- Called to free a Bignum value used in intermediate computations. In
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-- this implementation using the secondary stack, it does nothing at all,
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-- because we rely on Mark/Release, but it may be of use for some
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-- alternative implementation.
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function Normalize
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(X : Digit_Vector;
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Neg : Boolean := False) return Bignum;
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-- Given a digit vector and sign, allocate and construct a Bignum value.
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-- Note that X may have leading zeroes which must be removed, and if the
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-- result is zero, the sign is forced positive.
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---------
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-- Add --
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---------
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function Add
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(X, Y : Digit_Vector;
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X_Neg : Boolean;
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Y_Neg : Boolean) return Bignum
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is
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begin
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-- If signs are the same, we are doing an addition, it is convenient to
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-- ensure that the first operand is the longer of the two.
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if X_Neg = Y_Neg then
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if X'Last < Y'Last then
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return Add (X => Y, Y => X, X_Neg => Y_Neg, Y_Neg => X_Neg);
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-- Here signs are the same, and the first operand is the longer
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else
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pragma Assert (X_Neg = Y_Neg and then X'Last >= Y'Last);
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-- Do addition, putting result in Sum (allowing for carry)
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declare
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Sum : Digit_Vector (0 .. X'Last);
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RD : DD;
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begin
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RD := 0;
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for J in reverse 1 .. X'Last loop
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RD := RD + DD (X (J));
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if J >= 1 + (X'Last - Y'Last) then
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RD := RD + DD (Y (J - (X'Last - Y'Last)));
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end if;
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Sum (J) := LSD (RD);
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RD := RD / Base;
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end loop;
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Sum (0) := SD (RD);
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return Normalize (Sum, X_Neg);
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end;
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end if;
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-- Signs are different so really this is a subtraction, we want to make
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-- sure that the largest magnitude operand is the first one, and then
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-- the result will have the sign of the first operand.
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else
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declare
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CR : constant Compare_Result := Compare (X, Y, False, False);
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begin
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if CR = EQ then
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return Normalize (Zero_Data);
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elsif CR = LT then
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return Add (X => Y, Y => X, X_Neg => Y_Neg, Y_Neg => X_Neg);
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else
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pragma Assert (X_Neg /= Y_Neg and then CR = GT);
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-- Do subtraction, putting result in Diff
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declare
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Diff : Digit_Vector (1 .. X'Length);
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RD : DD;
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begin
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RD := 0;
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for J in reverse 1 .. X'Last loop
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RD := RD + DD (X (J));
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if J >= 1 + (X'Last - Y'Last) then
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RD := RD - DD (Y (J - (X'Last - Y'Last)));
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end if;
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Diff (J) := LSD (RD);
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RD := (if RD < Base then 0 else -1);
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end loop;
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return Normalize (Diff, X_Neg);
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end;
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end if;
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end;
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end if;
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end Add;
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---------------------
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-- Allocate_Bignum --
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---------------------
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function Allocate_Bignum (Len : Length) return Bignum is
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Addr : Address;
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begin
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-- Change the if False here to if True to get allocation on the heap
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-- instead of the secondary stack, which is convenient for debugging
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-- System.Bignum itself.
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if False then
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declare
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B : Bignum;
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begin
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B := new Bignum_Data'(Len, False, (others => 0));
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return B;
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end;
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-- Normal case of allocation on the secondary stack
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else
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-- Note: The approach used here is designed to avoid strict aliasing
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-- warnings that appeared previously using unchecked conversion.
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SS_Allocate (Addr, Storage_Offset (4 + 4 * Len));
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declare
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B : Bignum;
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for B'Address use Addr'Address;
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pragma Import (Ada, B);
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BD : Bignum_Data (Len);
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for BD'Address use Addr;
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pragma Import (Ada, BD);
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-- Expose a writable view of discriminant BD.Len so that we can
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-- initialize it. We need to use the exact layout of the record
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-- to ensure that the Length field has 24 bits as expected.
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type Bignum_Data_Header is record
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Len : Length;
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Neg : Boolean;
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end record;
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for Bignum_Data_Header use record
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Len at 0 range 0 .. 23;
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Neg at 3 range 0 .. 7;
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end record;
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BDH : Bignum_Data_Header;
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for BDH'Address use BD'Address;
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pragma Import (Ada, BDH);
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pragma Assert (BDH.Len'Size = BD.Len'Size);
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begin
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BDH.Len := Len;
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return B;
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end;
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end if;
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end Allocate_Bignum;
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-------------
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-- Big_Abs --
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-------------
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function Big_Abs (X : Bignum) return Bignum is
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begin
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return Normalize (X.D);
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end Big_Abs;
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-------------
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-- Big_Add --
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-------------
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function Big_Add (X, Y : Bignum) return Bignum is
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begin
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return Add (X.D, Y.D, X.Neg, Y.Neg);
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end Big_Add;
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-------------
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-- Big_Div --
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-------------
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-- This table is excerpted from RM 4.5.5(28-30) and shows how the result
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-- varies with the signs of the operands.
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-- A B A/B A B A/B
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--
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-- 10 5 2 -10 5 -2
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-- 11 5 2 -11 5 -2
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-- 12 5 2 -12 5 -2
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-- 13 5 2 -13 5 -2
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-- 14 5 2 -14 5 -2
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--
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-- A B A/B A B A/B
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--
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-- 10 -5 -2 -10 -5 2
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-- 11 -5 -2 -11 -5 2
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-- 12 -5 -2 -12 -5 2
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-- 13 -5 -2 -13 -5 2
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-- 14 -5 -2 -14 -5 2
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function Big_Div (X, Y : Bignum) return Bignum is
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Q, R : Bignum;
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begin
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Div_Rem (X, Y, Q, R, Discard_Remainder => True);
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Q.Neg := Q.Len > 0 and then (X.Neg xor Y.Neg);
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return Q;
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end Big_Div;
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-------------
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-- Big_Exp --
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-------------
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function Big_Exp (X, Y : Bignum) return Bignum is
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function "**" (X : Bignum; Y : SD) return Bignum;
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-- Internal routine where we know right operand is one word
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----------
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-- "**" --
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----------
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function "**" (X : Bignum; Y : SD) return Bignum is
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begin
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case Y is
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-- X ** 0 is 1
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when 0 =>
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return Normalize (One_Data);
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-- X ** 1 is X
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when 1 =>
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return Normalize (X.D);
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-- X ** 2 is X * X
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when 2 =>
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return Big_Mul (X, X);
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-- For X greater than 2, use the recursion
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-- X even, X ** Y = (X ** (Y/2)) ** 2;
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-- X odd, X ** Y = (X ** (Y/2)) ** 2 * X;
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when others =>
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declare
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XY2 : constant Bignum := X ** (Y / 2);
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XY2S : constant Bignum := Big_Mul (XY2, XY2);
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Res : Bignum;
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begin
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Free_Bignum (XY2);
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-- Raise storage error if intermediate value is getting too
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-- large, which we arbitrarily define as 200 words for now.
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if XY2S.Len > 200 then
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Free_Bignum (XY2S);
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raise Storage_Error with
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"exponentiation result is too large";
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end if;
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-- Otherwise take care of even/odd cases
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if (Y and 1) = 0 then
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return XY2S;
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else
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Res := Big_Mul (XY2S, X);
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Free_Bignum (XY2S);
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return Res;
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end if;
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end;
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end case;
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end "**";
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-- Start of processing for Big_Exp
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begin
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-- Error if right operand negative
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if Y.Neg then
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raise Constraint_Error with "exponentiation to negative power";
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-- X ** 0 is always 1 (including 0 ** 0, so do this test first)
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elsif Y.Len = 0 then
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return Normalize (One_Data);
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-- 0 ** X is always 0 (for X non-zero)
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elsif X.Len = 0 then
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return Normalize (Zero_Data);
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-- (+1) ** Y = 1
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-- (-1) ** Y = +/-1 depending on whether Y is even or odd
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elsif X.Len = 1 and then X.D (1) = 1 then
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return Normalize
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(X.D, Neg => X.Neg and then ((Y.D (Y.Len) and 1) = 1));
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-- If the absolute value of the base is greater than 1, then the
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-- exponent must not be bigger than one word, otherwise the result
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-- is ludicrously large, and we just signal Storage_Error right away.
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elsif Y.Len > 1 then
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raise Storage_Error with "exponentiation result is too large";
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-- Special case (+/-)2 ** K, where K is 1 .. 31 using a shift
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elsif X.Len = 1 and then X.D (1) = 2 and then Y.D (1) < 32 then
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declare
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D : constant Digit_Vector (1 .. 1) :=
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(1 => Shift_Left (SD'(1), Natural (Y.D (1))));
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begin
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return Normalize (D, X.Neg);
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end;
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-- Remaining cases have right operand of one word
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else
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return X ** Y.D (1);
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end if;
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end Big_Exp;
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------------
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-- Big_EQ --
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------------
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function Big_EQ (X, Y : Bignum) return Boolean is
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begin
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return Compare (X.D, Y.D, X.Neg, Y.Neg) = EQ;
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end Big_EQ;
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------------
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-- Big_GE --
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------------
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function Big_GE (X, Y : Bignum) return Boolean is
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begin
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return Compare (X.D, Y.D, X.Neg, Y.Neg) /= LT;
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end Big_GE;
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------------
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-- Big_GT --
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------------
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function Big_GT (X, Y : Bignum) return Boolean is
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begin
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return Compare (X.D, Y.D, X.Neg, Y.Neg) = GT;
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end Big_GT;
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------------
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-- Big_LE --
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------------
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function Big_LE (X, Y : Bignum) return Boolean is
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begin
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return Compare (X.D, Y.D, X.Neg, Y.Neg) /= GT;
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end Big_LE;
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------------
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-- Big_LT --
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------------
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function Big_LT (X, Y : Bignum) return Boolean is
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begin
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return Compare (X.D, Y.D, X.Neg, Y.Neg) = LT;
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end Big_LT;
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-------------
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-- Big_Mod --
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-------------
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-- This table is excerpted from RM 4.5.5(28-30) and shows how the result
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-- of Rem and Mod vary with the signs of the operands.
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-- A B A mod B A rem B A B A mod B A rem B
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-- 10 5 0 0 -10 5 0 0
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-- 11 5 1 1 -11 5 4 -1
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-- 12 5 2 2 -12 5 3 -2
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-- 13 5 3 3 -13 5 2 -3
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-- 14 5 4 4 -14 5 1 -4
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-- A B A mod B A rem B A B A mod B A rem B
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-- 10 -5 0 0 -10 -5 0 0
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-- 11 -5 -4 1 -11 -5 -1 -1
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-- 12 -5 -3 2 -12 -5 -2 -2
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-- 13 -5 -2 3 -13 -5 -3 -3
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-- 14 -5 -1 4 -14 -5 -4 -4
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function Big_Mod (X, Y : Bignum) return Bignum is
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Q, R : Bignum;
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begin
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-- If signs are same, result is same as Rem
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if X.Neg = Y.Neg then
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return Big_Rem (X, Y);
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-- Case where Mod is different
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else
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-- Do division
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Div_Rem (X, Y, Q, R, Discard_Quotient => True);
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-- Zero result is unchanged
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if R.Len = 0 then
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return R;
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-- Otherwise adjust result
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else
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declare
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T1 : constant Bignum := Big_Sub (Y, R);
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begin
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T1.Neg := Y.Neg;
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Free_Bignum (R);
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return T1;
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end;
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end if;
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end if;
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end Big_Mod;
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|
|
-------------
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-- Big_Mul --
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-------------
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function Big_Mul (X, Y : Bignum) return Bignum is
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Result : Digit_Vector (1 .. X.Len + Y.Len) := (others => 0);
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-- Accumulate result (max length of result is sum of operand lengths)
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L : Length;
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-- Current result digit
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D : DD;
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-- Result digit
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begin
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for J in 1 .. X.Len loop
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for K in 1 .. Y.Len loop
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L := Result'Last - (X.Len - J) - (Y.Len - K);
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D := DD (X.D (J)) * DD (Y.D (K)) + DD (Result (L));
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Result (L) := LSD (D);
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D := D / Base;
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-- D is carry which must be propagated
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while D /= 0 and then L >= 1 loop
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L := L - 1;
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D := D + DD (Result (L));
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Result (L) := LSD (D);
|
|
D := D / Base;
|
|
end loop;
|
|
|
|
-- Must not have a carry trying to extend max length
|
|
|
|
pragma Assert (D = 0);
|
|
end loop;
|
|
end loop;
|
|
|
|
-- Return result
|
|
|
|
return Normalize (Result, X.Neg xor Y.Neg);
|
|
end Big_Mul;
|
|
|
|
------------
|
|
-- Big_NE --
|
|
------------
|
|
|
|
function Big_NE (X, Y : Bignum) return Boolean is
|
|
begin
|
|
return Compare (X.D, Y.D, X.Neg, Y.Neg) /= EQ;
|
|
end Big_NE;
|
|
|
|
-------------
|
|
-- Big_Neg --
|
|
-------------
|
|
|
|
function Big_Neg (X : Bignum) return Bignum is
|
|
begin
|
|
return Normalize (X.D, not X.Neg);
|
|
end Big_Neg;
|
|
|
|
-------------
|
|
-- Big_Rem --
|
|
-------------
|
|
|
|
-- This table is excerpted from RM 4.5.5(28-30) and shows how the result
|
|
-- varies with the signs of the operands.
|
|
|
|
-- A B A rem B A B A rem B
|
|
|
|
-- 10 5 0 -10 5 0
|
|
-- 11 5 1 -11 5 -1
|
|
-- 12 5 2 -12 5 -2
|
|
-- 13 5 3 -13 5 -3
|
|
-- 14 5 4 -14 5 -4
|
|
|
|
-- A B A rem B A B A rem B
|
|
|
|
-- 10 -5 0 -10 -5 0
|
|
-- 11 -5 1 -11 -5 -1
|
|
-- 12 -5 2 -12 -5 -2
|
|
-- 13 -5 3 -13 -5 -3
|
|
-- 14 -5 4 -14 -5 -4
|
|
|
|
function Big_Rem (X, Y : Bignum) return Bignum is
|
|
Q, R : Bignum;
|
|
begin
|
|
Div_Rem (X, Y, Q, R, Discard_Quotient => True);
|
|
R.Neg := R.Len > 0 and then X.Neg;
|
|
return R;
|
|
end Big_Rem;
|
|
|
|
-------------
|
|
-- Big_Sub --
|
|
-------------
|
|
|
|
function Big_Sub (X, Y : Bignum) return Bignum is
|
|
begin
|
|
-- If right operand zero, return left operand (avoiding sharing)
|
|
|
|
if Y.Len = 0 then
|
|
return Normalize (X.D, X.Neg);
|
|
|
|
-- Otherwise add negative of right operand
|
|
|
|
else
|
|
return Add (X.D, Y.D, X.Neg, not Y.Neg);
|
|
end if;
|
|
end Big_Sub;
|
|
|
|
-------------
|
|
-- Compare --
|
|
-------------
|
|
|
|
function Compare
|
|
(X, Y : Digit_Vector;
|
|
X_Neg, Y_Neg : Boolean) return Compare_Result
|
|
is
|
|
begin
|
|
-- Signs are different, that's decisive, since 0 is always plus
|
|
|
|
if X_Neg /= Y_Neg then
|
|
return (if X_Neg then LT else GT);
|
|
|
|
-- Lengths are different, that's decisive since no leading zeroes
|
|
|
|
elsif X'Last /= Y'Last then
|
|
return (if (X'Last > Y'Last) xor X_Neg then GT else LT);
|
|
|
|
-- Need to compare data
|
|
|
|
else
|
|
for J in X'Range loop
|
|
if X (J) /= Y (J) then
|
|
return (if (X (J) > Y (J)) xor X_Neg then GT else LT);
|
|
end if;
|
|
end loop;
|
|
|
|
return EQ;
|
|
end if;
|
|
end Compare;
|
|
|
|
-------------
|
|
-- Div_Rem --
|
|
-------------
|
|
|
|
procedure Div_Rem
|
|
(X, Y : Bignum;
|
|
Quotient : out Bignum;
|
|
Remainder : out Bignum;
|
|
Discard_Quotient : Boolean := False;
|
|
Discard_Remainder : Boolean := False)
|
|
is
|
|
begin
|
|
-- Error if division by zero
|
|
|
|
if Y.Len = 0 then
|
|
raise Constraint_Error with "division by zero";
|
|
end if;
|
|
|
|
-- Handle simple cases with special tests
|
|
|
|
-- If X < Y then quotient is zero and remainder is X
|
|
|
|
if Compare (X.D, Y.D, False, False) = LT then
|
|
Remainder := Normalize (X.D);
|
|
Quotient := Normalize (Zero_Data);
|
|
return;
|
|
|
|
-- If both X and Y are less than 2**63-1, we can use Long_Long_Integer
|
|
-- arithmetic. Note it is good not to do an accurate range check against
|
|
-- Long_Long_Integer since -2**63 / -1 overflows.
|
|
|
|
elsif (X.Len <= 1 or else (X.Len = 2 and then X.D (1) < 2**31))
|
|
and then
|
|
(Y.Len <= 1 or else (Y.Len = 2 and then Y.D (1) < 2**31))
|
|
then
|
|
declare
|
|
A : constant LLI := abs (From_Bignum (X));
|
|
B : constant LLI := abs (From_Bignum (Y));
|
|
begin
|
|
Quotient := To_Bignum (A / B);
|
|
Remainder := To_Bignum (A rem B);
|
|
return;
|
|
end;
|
|
|
|
-- Easy case if divisor is one digit
|
|
|
|
elsif Y.Len = 1 then
|
|
declare
|
|
ND : DD;
|
|
Div : constant DD := DD (Y.D (1));
|
|
|
|
Result : Digit_Vector (1 .. X.Len);
|
|
Remdr : Digit_Vector (1 .. 1);
|
|
|
|
begin
|
|
ND := 0;
|
|
for J in 1 .. X.Len loop
|
|
ND := Base * ND + DD (X.D (J));
|
|
Result (J) := SD (ND / Div);
|
|
ND := ND rem Div;
|
|
end loop;
|
|
|
|
Quotient := Normalize (Result);
|
|
Remdr (1) := SD (ND);
|
|
Remainder := Normalize (Remdr);
|
|
return;
|
|
end;
|
|
end if;
|
|
|
|
-- The complex full multi-precision case. We will employ algorithm
|
|
-- D defined in the section "The Classical Algorithms" (sec. 4.3.1)
|
|
-- of Donald Knuth's "The Art of Computer Programming", Vol. 2, 2nd
|
|
-- edition. The terminology is adjusted for this section to match that
|
|
-- reference.
|
|
|
|
-- We are dividing X.Len digits of X (called u here) by Y.Len digits
|
|
-- of Y (called v here), developing the quotient and remainder. The
|
|
-- numbers are represented using Base, which was chosen so that we have
|
|
-- the operations of multiplying to single digits (SD) to form a double
|
|
-- digit (DD), and dividing a double digit (DD) by a single digit (SD)
|
|
-- to give a single digit quotient and a single digit remainder.
|
|
|
|
-- Algorithm D from Knuth
|
|
|
|
-- Comments here with square brackets are directly from Knuth
|
|
|
|
Algorithm_D : declare
|
|
|
|
-- The following lower case variables correspond exactly to the
|
|
-- terminology used in algorithm D.
|
|
|
|
m : constant Length := X.Len - Y.Len;
|
|
n : constant Length := Y.Len;
|
|
b : constant DD := Base;
|
|
|
|
u : Digit_Vector (0 .. m + n);
|
|
v : Digit_Vector (1 .. n);
|
|
q : Digit_Vector (0 .. m);
|
|
r : Digit_Vector (1 .. n);
|
|
|
|
u0 : SD renames u (0);
|
|
v1 : SD renames v (1);
|
|
v2 : SD renames v (2);
|
|
|
|
d : DD;
|
|
j : Length;
|
|
qhat : DD;
|
|
rhat : DD;
|
|
temp : DD;
|
|
|
|
begin
|
|
-- Initialize data of left and right operands
|
|
|
|
for J in 1 .. m + n loop
|
|
u (J) := X.D (J);
|
|
end loop;
|
|
|
|
for J in 1 .. n loop
|
|
v (J) := Y.D (J);
|
|
end loop;
|
|
|
|
-- [Division of nonnegative integers.] Given nonnegative integers u
|
|
-- = (ul,u2..um+n) and v = (v1,v2..vn), where v1 /= 0 and n > 1, we
|
|
-- form the quotient u / v = (q0,ql..qm) and the remainder u mod v =
|
|
-- (r1,r2..rn).
|
|
|
|
pragma Assert (v1 /= 0);
|
|
pragma Assert (n > 1);
|
|
|
|
-- Dl. [Normalize.] Set d = b/(vl + 1). Then set (u0,u1,u2..um+n)
|
|
-- equal to (u1,u2..um+n) times d, and set (v1,v2..vn) equal to
|
|
-- (v1,v2..vn) times d. Note the introduction of a new digit position
|
|
-- u0 at the left of u1; if d = 1 all we need to do in this step is
|
|
-- to set u0 = 0.
|
|
|
|
d := b / (DD (v1) + 1);
|
|
|
|
if d = 1 then
|
|
u0 := 0;
|
|
|
|
else
|
|
declare
|
|
Carry : DD;
|
|
Tmp : DD;
|
|
|
|
begin
|
|
-- Multiply Dividend (u) by d
|
|
|
|
Carry := 0;
|
|
for J in reverse 1 .. m + n loop
|
|
Tmp := DD (u (J)) * d + Carry;
|
|
u (J) := LSD (Tmp);
|
|
Carry := Tmp / Base;
|
|
end loop;
|
|
|
|
u0 := SD (Carry);
|
|
|
|
-- Multiply Divisor (v) by d
|
|
|
|
Carry := 0;
|
|
for J in reverse 1 .. n loop
|
|
Tmp := DD (v (J)) * d + Carry;
|
|
v (J) := LSD (Tmp);
|
|
Carry := Tmp / Base;
|
|
end loop;
|
|
|
|
pragma Assert (Carry = 0);
|
|
end;
|
|
end if;
|
|
|
|
-- D2. [Initialize j.] Set j = 0. The loop on j, steps D2 through D7,
|
|
-- will be essentially a division of (uj, uj+1..uj+n) by (v1,v2..vn)
|
|
-- to get a single quotient digit qj.
|
|
|
|
j := 0;
|
|
|
|
-- Loop through digits
|
|
|
|
loop
|
|
-- Note: In the original printing, step D3 was as follows:
|
|
|
|
-- D3. [Calculate qhat.] If uj = v1, set qhat to b-l; otherwise
|
|
-- set qhat to (uj,uj+1)/v1. Now test if v2 * qhat is greater than
|
|
-- (uj*b + uj+1 - qhat*v1)*b + uj+2. If so, decrease qhat by 1 and
|
|
-- repeat this test
|
|
|
|
-- This had a bug not discovered till 1995, see Vol 2 errata:
|
|
-- http://www-cs-faculty.stanford.edu/~uno/err2-2e.ps.gz. Under
|
|
-- rare circumstances the expression in the test could overflow.
|
|
-- This version was further corrected in 2005, see Vol 2 errata:
|
|
-- http://www-cs-faculty.stanford.edu/~uno/all2-pre.ps.gz.
|
|
-- The code below is the fixed version of this step.
|
|
|
|
-- D3. [Calculate qhat.] Set qhat to (uj,uj+1)/v1 and rhat to
|
|
-- to (uj,uj+1) mod v1.
|
|
|
|
temp := u (j) & u (j + 1);
|
|
qhat := temp / DD (v1);
|
|
rhat := temp mod DD (v1);
|
|
|
|
-- D3 (continued). Now test if qhat >= b or v2*qhat > (rhat,uj+2):
|
|
-- if so, decrease qhat by 1, increase rhat by v1, and repeat this
|
|
-- test if rhat < b. [The test on v2 determines at high speed
|
|
-- most of the cases in which the trial value qhat is one too
|
|
-- large, and eliminates all cases where qhat is two too large.]
|
|
|
|
while qhat >= b
|
|
or else DD (v2) * qhat > LSD (rhat) & u (j + 2)
|
|
loop
|
|
qhat := qhat - 1;
|
|
rhat := rhat + DD (v1);
|
|
exit when rhat >= b;
|
|
end loop;
|
|
|
|
-- D4. [Multiply and subtract.] Replace (uj,uj+1..uj+n) by
|
|
-- (uj,uj+1..uj+n) minus qhat times (v1,v2..vn). This step
|
|
-- consists of a simple multiplication by a one-place number,
|
|
-- combined with a subtraction.
|
|
|
|
-- The digits (uj,uj+1..uj+n) are always kept positive; if the
|
|
-- result of this step is actually negative then (uj,uj+1..uj+n)
|
|
-- is left as the true value plus b**(n+1), i.e. as the b's
|
|
-- complement of the true value, and a "borrow" to the left is
|
|
-- remembered.
|
|
|
|
declare
|
|
Borrow : SD;
|
|
Carry : DD;
|
|
Temp : DD;
|
|
|
|
Negative : Boolean;
|
|
-- Records if subtraction causes a negative result, requiring
|
|
-- an add back (case where qhat turned out to be 1 too large).
|
|
|
|
begin
|
|
Borrow := 0;
|
|
for K in reverse 1 .. n loop
|
|
Temp := qhat * DD (v (K)) + DD (Borrow);
|
|
Borrow := MSD (Temp);
|
|
|
|
if LSD (Temp) > u (j + K) then
|
|
Borrow := Borrow + 1;
|
|
end if;
|
|
|
|
u (j + K) := u (j + K) - LSD (Temp);
|
|
end loop;
|
|
|
|
Negative := u (j) < Borrow;
|
|
u (j) := u (j) - Borrow;
|
|
|
|
-- D5. [Test remainder.] Set qj = qhat. If the result of step
|
|
-- D4 was negative, we will do the add back step (step D6).
|
|
|
|
q (j) := LSD (qhat);
|
|
|
|
if Negative then
|
|
|
|
-- D6. [Add back.] Decrease qj by 1, and add (0,v1,v2..vn)
|
|
-- to (uj,uj+1,uj+2..uj+n). (A carry will occur to the left
|
|
-- of uj, and it is be ignored since it cancels with the
|
|
-- borrow that occurred in D4.)
|
|
|
|
q (j) := q (j) - 1;
|
|
|
|
Carry := 0;
|
|
for K in reverse 1 .. n loop
|
|
Temp := DD (v (K)) + DD (u (j + K)) + Carry;
|
|
u (j + K) := LSD (Temp);
|
|
Carry := Temp / Base;
|
|
end loop;
|
|
|
|
u (j) := u (j) + SD (Carry);
|
|
end if;
|
|
end;
|
|
|
|
-- D7. [Loop on j.] Increase j by one. Now if j <= m, go back to
|
|
-- D3 (the start of the loop on j).
|
|
|
|
j := j + 1;
|
|
exit when not (j <= m);
|
|
end loop;
|
|
|
|
-- D8. [Unnormalize.] Now (qo,ql..qm) is the desired quotient, and
|
|
-- the desired remainder may be obtained by dividing (um+1..um+n)
|
|
-- by d.
|
|
|
|
if not Discard_Quotient then
|
|
Quotient := Normalize (q);
|
|
end if;
|
|
|
|
if not Discard_Remainder then
|
|
declare
|
|
Remdr : DD;
|
|
|
|
begin
|
|
Remdr := 0;
|
|
for K in 1 .. n loop
|
|
Remdr := Base * Remdr + DD (u (m + K));
|
|
r (K) := SD (Remdr / d);
|
|
Remdr := Remdr rem d;
|
|
end loop;
|
|
|
|
pragma Assert (Remdr = 0);
|
|
end;
|
|
|
|
Remainder := Normalize (r);
|
|
end if;
|
|
end Algorithm_D;
|
|
end Div_Rem;
|
|
|
|
-----------------
|
|
-- From_Bignum --
|
|
-----------------
|
|
|
|
function From_Bignum (X : Bignum) return Long_Long_Integer is
|
|
begin
|
|
if X.Len = 0 then
|
|
return 0;
|
|
|
|
elsif X.Len = 1 then
|
|
return (if X.Neg then -LLI (X.D (1)) else LLI (X.D (1)));
|
|
|
|
elsif X.Len = 2 then
|
|
declare
|
|
Mag : constant DD := X.D (1) & X.D (2);
|
|
begin
|
|
if X.Neg and then Mag <= 2 ** 63 then
|
|
return -LLI (Mag);
|
|
elsif Mag < 2 ** 63 then
|
|
return LLI (Mag);
|
|
end if;
|
|
end;
|
|
end if;
|
|
|
|
raise Constraint_Error with "expression value out of range";
|
|
end From_Bignum;
|
|
|
|
-------------------------
|
|
-- Bignum_In_LLI_Range --
|
|
-------------------------
|
|
|
|
function Bignum_In_LLI_Range (X : Bignum) return Boolean is
|
|
begin
|
|
-- If length is 0 or 1, definitely fits
|
|
|
|
if X.Len <= 1 then
|
|
return True;
|
|
|
|
-- If length is greater than 2, definitely does not fit
|
|
|
|
elsif X.Len > 2 then
|
|
return False;
|
|
|
|
-- Length is 2, more tests needed
|
|
|
|
else
|
|
declare
|
|
Mag : constant DD := X.D (1) & X.D (2);
|
|
begin
|
|
return Mag < 2 ** 63 or else (X.Neg and then Mag = 2 ** 63);
|
|
end;
|
|
end if;
|
|
end Bignum_In_LLI_Range;
|
|
|
|
---------------
|
|
-- Normalize --
|
|
---------------
|
|
|
|
function Normalize
|
|
(X : Digit_Vector;
|
|
Neg : Boolean := False) return Bignum
|
|
is
|
|
B : Bignum;
|
|
J : Length;
|
|
|
|
begin
|
|
J := X'First;
|
|
while J <= X'Last and then X (J) = 0 loop
|
|
J := J + 1;
|
|
end loop;
|
|
|
|
B := Allocate_Bignum (X'Last - J + 1);
|
|
B.Neg := B.Len > 0 and then Neg;
|
|
B.D := X (J .. X'Last);
|
|
return B;
|
|
end Normalize;
|
|
|
|
---------------
|
|
-- To_Bignum --
|
|
---------------
|
|
|
|
function To_Bignum (X : Long_Long_Integer) return Bignum is
|
|
R : Bignum;
|
|
|
|
begin
|
|
if X = 0 then
|
|
R := Allocate_Bignum (0);
|
|
|
|
-- One word result
|
|
|
|
elsif X in -(2 ** 32 - 1) .. +(2 ** 32 - 1) then
|
|
R := Allocate_Bignum (1);
|
|
R.D (1) := SD (abs (X));
|
|
|
|
-- Largest negative number annoyance
|
|
|
|
elsif X = Long_Long_Integer'First then
|
|
R := Allocate_Bignum (2);
|
|
R.D (1) := 2 ** 31;
|
|
R.D (2) := 0;
|
|
|
|
-- Normal two word case
|
|
|
|
else
|
|
R := Allocate_Bignum (2);
|
|
R.D (2) := SD (abs (X) mod Base);
|
|
R.D (1) := SD (abs (X) / Base);
|
|
end if;
|
|
|
|
R.Neg := X < 0;
|
|
return R;
|
|
end To_Bignum;
|
|
|
|
end System.Bignums;
|