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CodeBlocksPortable/MinGW/lib/gcc/mingw32/6.3.0/adainclude/s-bignum.adb

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Ada

------------------------------------------------------------------------------
-- --
-- GNAT COMPILER COMPONENTS --
-- --
-- S Y S T E M . B I G N U M S --
-- --
-- B o d y --
-- --
-- Copyright (C) 2012-2015, Free Software Foundation, Inc. --
-- --
-- GNAT is free software; you can redistribute it and/or modify it under --
-- terms of the GNU General Public License as published by the Free Soft- --
-- ware Foundation; either version 3, or (at your option) any later ver- --
-- sion. GNAT is distributed in the hope that it will be useful, but WITH- --
-- OUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY --
-- or FITNESS FOR A PARTICULAR PURPOSE. --
-- --
-- As a special exception under Section 7 of GPL version 3, you are granted --
-- additional permissions described in the GCC Runtime Library Exception, --
-- version 3.1, as published by the Free Software Foundation. --
-- --
-- You should have received a copy of the GNU General Public License and --
-- a copy of the GCC Runtime Library Exception along with this program; --
-- see the files COPYING3 and COPYING.RUNTIME respectively. If not, see --
-- <http://www.gnu.org/licenses/>. --
-- --
-- GNAT was originally developed by the GNAT team at New York University. --
-- Extensive contributions were provided by Ada Core Technologies Inc. --
-- --
------------------------------------------------------------------------------
-- This package provides arbitrary precision signed integer arithmetic for
-- use in computing intermediate values in expressions for the case where
-- pragma Overflow_Check (Eliminate) is in effect.
with System; use System;
with System.Secondary_Stack; use System.Secondary_Stack;
with System.Storage_Elements; use System.Storage_Elements;
package body System.Bignums is
use Interfaces;
-- So that operations on Unsigned_32 are available
type DD is mod Base ** 2;
-- Double length digit used for intermediate computations
function MSD (X : DD) return SD is (SD (X / Base));
function LSD (X : DD) return SD is (SD (X mod Base));
-- Most significant and least significant digit of double digit value
function "&" (X, Y : SD) return DD is (DD (X) * Base + DD (Y));
-- Compose double digit value from two single digit values
subtype LLI is Long_Long_Integer;
One_Data : constant Digit_Vector (1 .. 1) := (1 => 1);
-- Constant one
Zero_Data : constant Digit_Vector (1 .. 0) := (1 .. 0 => 0);
-- Constant zero
-----------------------
-- Local Subprograms --
-----------------------
function Add
(X, Y : Digit_Vector;
X_Neg : Boolean;
Y_Neg : Boolean) return Bignum
with
Pre => X'First = 1 and then Y'First = 1;
-- This procedure adds two signed numbers returning the Sum, it is used
-- for both addition and subtraction. The value computed is X + Y, with
-- X_Neg and Y_Neg giving the signs of the operands.
function Allocate_Bignum (Len : Length) return Bignum with
Post => Allocate_Bignum'Result.Len = Len;
-- Allocate Bignum value of indicated length on secondary stack. On return
-- the Neg and D fields are left uninitialized.
type Compare_Result is (LT, EQ, GT);
-- Indicates result of comparison in following call
function Compare
(X, Y : Digit_Vector;
X_Neg, Y_Neg : Boolean) return Compare_Result
with
Pre => X'First = 1 and then Y'First = 1;
-- Compare (X with sign X_Neg) with (Y with sign Y_Neg), and return the
-- result of the signed comparison.
procedure Div_Rem
(X, Y : Bignum;
Quotient : out Bignum;
Remainder : out Bignum;
Discard_Quotient : Boolean := False;
Discard_Remainder : Boolean := False);
-- Returns the Quotient and Remainder from dividing abs (X) by abs (Y). The
-- values of X and Y are not modified. If Discard_Quotient is True, then
-- Quotient is undefined on return, and if Discard_Remainder is True, then
-- Remainder is undefined on return. Service routine for Big_Div/Rem/Mod.
procedure Free_Bignum (X : Bignum) is null;
-- Called to free a Bignum value used in intermediate computations. In
-- this implementation using the secondary stack, it does nothing at all,
-- because we rely on Mark/Release, but it may be of use for some
-- alternative implementation.
function Normalize
(X : Digit_Vector;
Neg : Boolean := False) return Bignum;
-- Given a digit vector and sign, allocate and construct a Bignum value.
-- Note that X may have leading zeroes which must be removed, and if the
-- result is zero, the sign is forced positive.
---------
-- Add --
---------
function Add
(X, Y : Digit_Vector;
X_Neg : Boolean;
Y_Neg : Boolean) return Bignum
is
begin
-- If signs are the same, we are doing an addition, it is convenient to
-- ensure that the first operand is the longer of the two.
if X_Neg = Y_Neg then
if X'Last < Y'Last then
return Add (X => Y, Y => X, X_Neg => Y_Neg, Y_Neg => X_Neg);
-- Here signs are the same, and the first operand is the longer
else
pragma Assert (X_Neg = Y_Neg and then X'Last >= Y'Last);
-- Do addition, putting result in Sum (allowing for carry)
declare
Sum : Digit_Vector (0 .. X'Last);
RD : DD;
begin
RD := 0;
for J in reverse 1 .. X'Last loop
RD := RD + DD (X (J));
if J >= 1 + (X'Last - Y'Last) then
RD := RD + DD (Y (J - (X'Last - Y'Last)));
end if;
Sum (J) := LSD (RD);
RD := RD / Base;
end loop;
Sum (0) := SD (RD);
return Normalize (Sum, X_Neg);
end;
end if;
-- Signs are different so really this is a subtraction, we want to make
-- sure that the largest magnitude operand is the first one, and then
-- the result will have the sign of the first operand.
else
declare
CR : constant Compare_Result := Compare (X, Y, False, False);
begin
if CR = EQ then
return Normalize (Zero_Data);
elsif CR = LT then
return Add (X => Y, Y => X, X_Neg => Y_Neg, Y_Neg => X_Neg);
else
pragma Assert (X_Neg /= Y_Neg and then CR = GT);
-- Do subtraction, putting result in Diff
declare
Diff : Digit_Vector (1 .. X'Length);
RD : DD;
begin
RD := 0;
for J in reverse 1 .. X'Last loop
RD := RD + DD (X (J));
if J >= 1 + (X'Last - Y'Last) then
RD := RD - DD (Y (J - (X'Last - Y'Last)));
end if;
Diff (J) := LSD (RD);
RD := (if RD < Base then 0 else -1);
end loop;
return Normalize (Diff, X_Neg);
end;
end if;
end;
end if;
end Add;
---------------------
-- Allocate_Bignum --
---------------------
function Allocate_Bignum (Len : Length) return Bignum is
Addr : Address;
begin
-- Change the if False here to if True to get allocation on the heap
-- instead of the secondary stack, which is convenient for debugging
-- System.Bignum itself.
if False then
declare
B : Bignum;
begin
B := new Bignum_Data'(Len, False, (others => 0));
return B;
end;
-- Normal case of allocation on the secondary stack
else
-- Note: The approach used here is designed to avoid strict aliasing
-- warnings that appeared previously using unchecked conversion.
SS_Allocate (Addr, Storage_Offset (4 + 4 * Len));
declare
B : Bignum;
for B'Address use Addr'Address;
pragma Import (Ada, B);
BD : Bignum_Data (Len);
for BD'Address use Addr;
pragma Import (Ada, BD);
-- Expose a writable view of discriminant BD.Len so that we can
-- initialize it. We need to use the exact layout of the record
-- to ensure that the Length field has 24 bits as expected.
type Bignum_Data_Header is record
Len : Length;
Neg : Boolean;
end record;
for Bignum_Data_Header use record
Len at 0 range 0 .. 23;
Neg at 3 range 0 .. 7;
end record;
BDH : Bignum_Data_Header;
for BDH'Address use BD'Address;
pragma Import (Ada, BDH);
pragma Assert (BDH.Len'Size = BD.Len'Size);
begin
BDH.Len := Len;
return B;
end;
end if;
end Allocate_Bignum;
-------------
-- Big_Abs --
-------------
function Big_Abs (X : Bignum) return Bignum is
begin
return Normalize (X.D);
end Big_Abs;
-------------
-- Big_Add --
-------------
function Big_Add (X, Y : Bignum) return Bignum is
begin
return Add (X.D, Y.D, X.Neg, Y.Neg);
end Big_Add;
-------------
-- Big_Div --
-------------
-- 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/B A B A/B
--
-- 10 5 2 -10 5 -2
-- 11 5 2 -11 5 -2
-- 12 5 2 -12 5 -2
-- 13 5 2 -13 5 -2
-- 14 5 2 -14 5 -2
--
-- A B A/B A B A/B
--
-- 10 -5 -2 -10 -5 2
-- 11 -5 -2 -11 -5 2
-- 12 -5 -2 -12 -5 2
-- 13 -5 -2 -13 -5 2
-- 14 -5 -2 -14 -5 2
function Big_Div (X, Y : Bignum) return Bignum is
Q, R : Bignum;
begin
Div_Rem (X, Y, Q, R, Discard_Remainder => True);
Q.Neg := Q.Len > 0 and then (X.Neg xor Y.Neg);
return Q;
end Big_Div;
-------------
-- Big_Exp --
-------------
function Big_Exp (X, Y : Bignum) return Bignum is
function "**" (X : Bignum; Y : SD) return Bignum;
-- Internal routine where we know right operand is one word
----------
-- "**" --
----------
function "**" (X : Bignum; Y : SD) return Bignum is
begin
case Y is
-- X ** 0 is 1
when 0 =>
return Normalize (One_Data);
-- X ** 1 is X
when 1 =>
return Normalize (X.D);
-- X ** 2 is X * X
when 2 =>
return Big_Mul (X, X);
-- For X greater than 2, use the recursion
-- X even, X ** Y = (X ** (Y/2)) ** 2;
-- X odd, X ** Y = (X ** (Y/2)) ** 2 * X;
when others =>
declare
XY2 : constant Bignum := X ** (Y / 2);
XY2S : constant Bignum := Big_Mul (XY2, XY2);
Res : Bignum;
begin
Free_Bignum (XY2);
-- Raise storage error if intermediate value is getting too
-- large, which we arbitrarily define as 200 words for now.
if XY2S.Len > 200 then
Free_Bignum (XY2S);
raise Storage_Error with
"exponentiation result is too large";
end if;
-- Otherwise take care of even/odd cases
if (Y and 1) = 0 then
return XY2S;
else
Res := Big_Mul (XY2S, X);
Free_Bignum (XY2S);
return Res;
end if;
end;
end case;
end "**";
-- Start of processing for Big_Exp
begin
-- Error if right operand negative
if Y.Neg then
raise Constraint_Error with "exponentiation to negative power";
-- X ** 0 is always 1 (including 0 ** 0, so do this test first)
elsif Y.Len = 0 then
return Normalize (One_Data);
-- 0 ** X is always 0 (for X non-zero)
elsif X.Len = 0 then
return Normalize (Zero_Data);
-- (+1) ** Y = 1
-- (-1) ** Y = +/-1 depending on whether Y is even or odd
elsif X.Len = 1 and then X.D (1) = 1 then
return Normalize
(X.D, Neg => X.Neg and then ((Y.D (Y.Len) and 1) = 1));
-- If the absolute value of the base is greater than 1, then the
-- exponent must not be bigger than one word, otherwise the result
-- is ludicrously large, and we just signal Storage_Error right away.
elsif Y.Len > 1 then
raise Storage_Error with "exponentiation result is too large";
-- Special case (+/-)2 ** K, where K is 1 .. 31 using a shift
elsif X.Len = 1 and then X.D (1) = 2 and then Y.D (1) < 32 then
declare
D : constant Digit_Vector (1 .. 1) :=
(1 => Shift_Left (SD'(1), Natural (Y.D (1))));
begin
return Normalize (D, X.Neg);
end;
-- Remaining cases have right operand of one word
else
return X ** Y.D (1);
end if;
end Big_Exp;
------------
-- Big_EQ --
------------
function Big_EQ (X, Y : Bignum) return Boolean is
begin
return Compare (X.D, Y.D, X.Neg, Y.Neg) = EQ;
end Big_EQ;
------------
-- Big_GE --
------------
function Big_GE (X, Y : Bignum) return Boolean is
begin
return Compare (X.D, Y.D, X.Neg, Y.Neg) /= LT;
end Big_GE;
------------
-- Big_GT --
------------
function Big_GT (X, Y : Bignum) return Boolean is
begin
return Compare (X.D, Y.D, X.Neg, Y.Neg) = GT;
end Big_GT;
------------
-- Big_LE --
------------
function Big_LE (X, Y : Bignum) return Boolean is
begin
return Compare (X.D, Y.D, X.Neg, Y.Neg) /= GT;
end Big_LE;
------------
-- Big_LT --
------------
function Big_LT (X, Y : Bignum) return Boolean is
begin
return Compare (X.D, Y.D, X.Neg, Y.Neg) = LT;
end Big_LT;
-------------
-- Big_Mod --
-------------
-- This table is excerpted from RM 4.5.5(28-30) and shows how the result
-- of Rem and Mod vary with the signs of the operands.
-- A B A mod B A rem B A B A mod B A rem B
-- 10 5 0 0 -10 5 0 0
-- 11 5 1 1 -11 5 4 -1
-- 12 5 2 2 -12 5 3 -2
-- 13 5 3 3 -13 5 2 -3
-- 14 5 4 4 -14 5 1 -4
-- A B A mod B A rem B A B A mod B A rem B
-- 10 -5 0 0 -10 -5 0 0
-- 11 -5 -4 1 -11 -5 -1 -1
-- 12 -5 -3 2 -12 -5 -2 -2
-- 13 -5 -2 3 -13 -5 -3 -3
-- 14 -5 -1 4 -14 -5 -4 -4
function Big_Mod (X, Y : Bignum) return Bignum is
Q, R : Bignum;
begin
-- If signs are same, result is same as Rem
if X.Neg = Y.Neg then
return Big_Rem (X, Y);
-- Case where Mod is different
else
-- Do division
Div_Rem (X, Y, Q, R, Discard_Quotient => True);
-- Zero result is unchanged
if R.Len = 0 then
return R;
-- Otherwise adjust result
else
declare
T1 : constant Bignum := Big_Sub (Y, R);
begin
T1.Neg := Y.Neg;
Free_Bignum (R);
return T1;
end;
end if;
end if;
end Big_Mod;
-------------
-- Big_Mul --
-------------
function Big_Mul (X, Y : Bignum) return Bignum is
Result : Digit_Vector (1 .. X.Len + Y.Len) := (others => 0);
-- Accumulate result (max length of result is sum of operand lengths)
L : Length;
-- Current result digit
D : DD;
-- Result digit
begin
for J in 1 .. X.Len loop
for K in 1 .. Y.Len loop
L := Result'Last - (X.Len - J) - (Y.Len - K);
D := DD (X.D (J)) * DD (Y.D (K)) + DD (Result (L));
Result (L) := LSD (D);
D := D / Base;
-- D is carry which must be propagated
while D /= 0 and then L >= 1 loop
L := L - 1;
D := D + DD (Result (L));
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;