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274 lines
7 KiB
C
274 lines
7 KiB
C
/* div.c: bcmath library file. */
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/*
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Copyright (C) 1991, 1992, 1993, 1994, 1997 Free Software Foundation, Inc.
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Copyright (C) 2000 Philip A. Nelson
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This library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2 of the License, or (at your option) any later version.
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This library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details. (COPYING.LIB)
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You should have received a copy of the GNU Lesser General Public
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License along with this library; if not, write to:
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The Free Software Foundation, Inc.
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59 Temple Place, Suite 330
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Boston, MA 02111-1307 USA.
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You may contact the author by:
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e-mail: philnelson@acm.org
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us-mail: Philip A. Nelson
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Computer Science Department, 9062
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Western Washington University
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Bellingham, WA 98226-9062
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*************************************************************************/
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#include <config.h>
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#include <stdio.h>
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#include <assert.h>
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#include <stdlib.h>
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#include <ctype.h>
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#include <stdarg.h>
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#include "bcmath.h"
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#include "private.h"
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/* Some utility routines for the divide: First a one digit multiply.
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NUM (with SIZE digits) is multiplied by DIGIT and the result is
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placed into RESULT. It is written so that NUM and RESULT can be
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the same pointers. */
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static void
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_one_mult (num, size, digit, result)
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unsigned char *num;
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int size, digit;
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unsigned char *result;
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{
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int carry, value;
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unsigned char *nptr, *rptr;
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if (digit == 0)
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memset (result, 0, size);
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else
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{
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if (digit == 1)
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memcpy (result, num, size);
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else
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{
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/* Initialize */
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nptr = (unsigned char *) (num+size-1);
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rptr = (unsigned char *) (result+size-1);
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carry = 0;
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while (size-- > 0)
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{
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value = *nptr-- * digit + carry;
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*rptr-- = value % BASE;
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carry = value / BASE;
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}
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if (carry != 0) *rptr = carry;
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}
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}
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}
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/* The full division routine. This computes N1 / N2. It returns
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0 if the division is ok and the result is in QUOT. The number of
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digits after the decimal point is SCALE. It returns -1 if division
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by zero is tried. The algorithm is found in Knuth Vol 2. p237. */
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int
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bc_divide (bc_num n1, bc_num n2, bc_num *quot, int scale)
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{
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bc_num qval;
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unsigned char *num1, *num2;
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unsigned char *ptr1, *ptr2, *n2ptr, *qptr;
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int scale1, val;
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unsigned int len1, len2, scale2, qdigits, extra, count;
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unsigned int qdig, qguess, borrow, carry;
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unsigned char *mval;
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char zero;
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unsigned int norm;
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/* Test for divide by zero. */
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if (bc_is_zero (n2)) return -1;
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/* Test for divide by 1. If it is we must truncate. */
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if (n2->n_scale == 0)
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{
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if (n2->n_len == 1 && *n2->n_value == 1)
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{
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qval = bc_new_num (n1->n_len, scale);
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qval->n_sign = (n1->n_sign == n2->n_sign ? PLUS : MINUS);
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memset (&qval->n_value[n1->n_len],0,scale);
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memcpy (qval->n_value, n1->n_value,
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n1->n_len + MIN(n1->n_scale,scale));
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bc_free_num (quot);
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*quot = qval;
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}
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}
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/* Set up the divide. Move the decimal point on n1 by n2's scale.
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Remember, zeros on the end of num2 are wasted effort for dividing. */
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scale2 = n2->n_scale;
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n2ptr = (unsigned char *) n2->n_value+n2->n_len+scale2-1;
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while ((scale2 > 0) && (*n2ptr-- == 0)) scale2--;
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len1 = n1->n_len + scale2;
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scale1 = n1->n_scale - scale2;
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if (scale1 < scale)
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extra = scale - scale1;
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else
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extra = 0;
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num1 = (unsigned char *) safe_emalloc (1, n1->n_len+n1->n_scale, extra+2);
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if (num1 == NULL) bc_out_of_memory();
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memset (num1, 0, n1->n_len+n1->n_scale+extra+2);
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memcpy (num1+1, n1->n_value, n1->n_len+n1->n_scale);
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len2 = n2->n_len + scale2;
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num2 = (unsigned char *) safe_emalloc (1, len2, 1);
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if (num2 == NULL) bc_out_of_memory();
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memcpy (num2, n2->n_value, len2);
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*(num2+len2) = 0;
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n2ptr = num2;
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while (*n2ptr == 0)
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{
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n2ptr++;
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len2--;
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}
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/* Calculate the number of quotient digits. */
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if (len2 > len1+scale)
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{
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qdigits = scale+1;
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zero = TRUE;
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}
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else
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{
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zero = FALSE;
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if (len2>len1)
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qdigits = scale+1; /* One for the zero integer part. */
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else
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qdigits = len1-len2+scale+1;
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}
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/* Allocate and zero the storage for the quotient. */
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qval = bc_new_num (qdigits-scale,scale);
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memset (qval->n_value, 0, qdigits);
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/* Allocate storage for the temporary storage mval. */
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mval = (unsigned char *) safe_emalloc (1, len2, 1);
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if (mval == NULL) bc_out_of_memory ();
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/* Now for the full divide algorithm. */
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if (!zero)
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{
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/* Normalize */
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norm = 10 / ((int)*n2ptr + 1);
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if (norm != 1)
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{
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_one_mult (num1, len1+scale1+extra+1, norm, num1);
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_one_mult (n2ptr, len2, norm, n2ptr);
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}
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/* Initialize divide loop. */
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qdig = 0;
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if (len2 > len1)
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qptr = (unsigned char *) qval->n_value+len2-len1;
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else
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qptr = (unsigned char *) qval->n_value;
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/* Loop */
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while (qdig <= len1+scale-len2)
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{
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/* Calculate the quotient digit guess. */
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if (*n2ptr == num1[qdig])
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qguess = 9;
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else
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qguess = (num1[qdig]*10 + num1[qdig+1]) / *n2ptr;
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/* Test qguess. */
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if (n2ptr[1]*qguess >
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(num1[qdig]*10 + num1[qdig+1] - *n2ptr*qguess)*10
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+ num1[qdig+2])
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{
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qguess--;
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/* And again. */
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if (n2ptr[1]*qguess >
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(num1[qdig]*10 + num1[qdig+1] - *n2ptr*qguess)*10
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+ num1[qdig+2])
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qguess--;
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}
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/* Multiply and subtract. */
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borrow = 0;
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if (qguess != 0)
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{
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*mval = 0;
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_one_mult (n2ptr, len2, qguess, mval+1);
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ptr1 = (unsigned char *) num1+qdig+len2;
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ptr2 = (unsigned char *) mval+len2;
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for (count = 0; count < len2+1; count++)
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{
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val = (int) *ptr1 - (int) *ptr2-- - borrow;
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if (val < 0)
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{
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val += 10;
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borrow = 1;
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}
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else
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borrow = 0;
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*ptr1-- = val;
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}
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}
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/* Test for negative result. */
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if (borrow == 1)
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{
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qguess--;
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ptr1 = (unsigned char *) num1+qdig+len2;
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ptr2 = (unsigned char *) n2ptr+len2-1;
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carry = 0;
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for (count = 0; count < len2; count++)
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{
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val = (int) *ptr1 + (int) *ptr2-- + carry;
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if (val > 9)
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{
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val -= 10;
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carry = 1;
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}
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else
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carry = 0;
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*ptr1-- = val;
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}
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if (carry == 1) *ptr1 = (*ptr1 + 1) % 10;
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}
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/* We now know the quotient digit. */
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*qptr++ = qguess;
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qdig++;
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}
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}
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/* Clean up and return the number. */
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qval->n_sign = ( n1->n_sign == n2->n_sign ? PLUS : MINUS );
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if (bc_is_zero (qval)) qval->n_sign = PLUS;
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_bc_rm_leading_zeros (qval);
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bc_free_num (quot);
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*quot = qval;
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/* Clean up temporary storage. */
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efree (mval);
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efree (num1);
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efree (num2);
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return 0; /* Everything is OK. */
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}
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