📄 bn_div.c
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/************************************************************
Copyright (C), 2004, Aerospace Information. Co., Ltd.
FileName: bn_div.cpp
Author: LSX Version : 1.0 Date:2004/9/16
Description: 大数除法运算,根据OpenSSL源码改编
Version: 1.0
Function List:
1. bn_div 两个大整数做除法运算
History:
<author> <time> <version > <desc>
LSX 2004/9/16 1.0 加入注释
************************************************************/
#include <stdio.h>
//#include <openssl/bn.h>
#include "bn_lcl.h"
/* The old slow way */
#if 0
/**************************************************************************************
Function: BN_div(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, const BIGNUM *d,
BN_CTX *ctx)
Description: 两个大整数做除法运算
Calls: bn_check_top, BN_is_zero, BN_ucmp, BN_copy, BN_num_bits, BN_zero,
BN_wexpand, BN_lshift, BN_lshift1, BN_usub, BN_rshift1,
BN_CTX_start, BN_CTX_get, BN_CTX_end
Called by:
Input: 两个大整数m和d, d不为0
Output: dv = m / d
rem = m % d
Return: 1, if 正确相除
0, otherwise
Others:
**************************************************************************************/
int BN_div(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, const BIGNUM *d, BN_CTX *ctx)
{
int i,nm,nd;
int ret = 0;
BIGNUM *D;
bn_check_top(m);
bn_check_top(d);
if (BN_is_zero(d))
{
fprintf(stderr,"BN_div by zero error!\n");
return(0);
}
if (BN_ucmp(m,d) < 0) /*m<d, we can do directly*/
{
if (rem != NULL)
{
if (BN_copy(rem,m) == NULL)
{
return(0);
}
}
if (dv != NULL) BN_zero(dv);
return(1);
}
BN_CTX_start(ctx);
D = BN_CTX_get(ctx);
if (dv == NULL)
{
dv = BN_CTX_get(ctx);
}
if (rem == NULL)
{
rem = BN_CTX_get(ctx);
}
if (D == NULL || dv == NULL || rem == NULL)
{
goto end;
}
nd=BN_num_bits(d);
nm=BN_num_bits(m);
if (BN_copy(D,d) == NULL)
{
goto end;
}
if (BN_copy(rem,m) == NULL)
{
goto end;
}
/* The next 2 are needed so we can do a dv->d[0]|=1 later
* since BN_lshift1 will only work once there is a value :-) */
BN_zero(dv);
bn_wexpand(dv,1);
dv->top=1;
if (!BN_lshift(D,D,nm-nd)) /*使得除数与被除数的bits一样*/
{
goto end;
}
for (i=nm-nd; i>=0; i--)
{
if (!BN_lshift1(dv,dv))
{
goto end;
}
if (BN_ucmp(rem,D) >= 0)
{
dv->d[0]|=1;
if (!BN_usub(rem,rem,D))
{
goto end;
}
}
if (!BN_rshift1(D,D))
{
goto end;
}
}
rem->neg=BN_is_zero(rem)?0:m->neg;
dv->neg=m->neg^d->neg;
ret = 1;
end:
BN_CTX_end(ctx);
return(ret);
}
#else
/* BN_div computes dv := num / divisor, rounding towards zero, and sets up
* rm such that dv*divisor + rm = num holds.
* Thus:
* dv->neg == num->neg ^ divisor->neg (unless the result is zero)
* rm->neg == num->neg (unless the remainder is zero)
* If 'dv' or 'rm' is NULL, the respective value is not returned.
*/
/**************************************************************************************
Function: BN_div(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, const BIGNUM *d,
BN_CTX *ctx)
Description: 两个大整数做除法运算
Calls: bn_check_top, BN_is_zero, BN_ucmp, BN_copy, BN_num_bits, BN_zero,
BN_wexpand, BN_lshift, BN_lshift1, BN_usub, BN_rshift1,
BN_CTX_start, BN_CTX_get, BN_CTX_end
Called by:
Input: 两个大整数m和d, d不为0
Output: dv = m / d
rem = m % d
Return: 1, if 正确相除
0, otherwise
Others:
**************************************************************************************/
int BN_div(BIGNUM *dv, BIGNUM *rm, const BIGNUM *num, const BIGNUM *divisor, BN_CTX *ctx)
{
int norm_shift,i,j,loop;
BIGNUM *tmp,wnum,*snum,*sdiv,*res;
BN_ULONG *resp,*wnump;
BN_ULONG d0,d1;
int num_n,div_n;
bn_check_top(num);
bn_check_top(divisor);
if (BN_is_zero(divisor))
{
fprintf(stderr,"BN_div by zero error!\n");
return(0);
}
if (BN_ucmp(num,divisor) < 0)
{
if (rm != NULL)
{
if (BN_copy(rm,num) == NULL)
{
return(0);
}
}
if (dv != NULL)
{
BN_zero(dv);
}
return(1);
}
BN_CTX_start(ctx);
tmp=BN_CTX_get(ctx);
snum=BN_CTX_get(ctx);
sdiv=BN_CTX_get(ctx);
if (dv == NULL)
{
res=BN_CTX_get(ctx);
}
else
{
res=dv;
}
if (sdiv == NULL || res == NULL)
{
goto err;
}
tmp->neg=0;
/* First we normalise the numbers */
norm_shift=BN_BITS2-((BN_num_bits(divisor))%BN_BITS2);
/* add by ZQS,no need normalist! */
if (norm_shift==BN_BITS2)
{
norm_shift=0;
}
if (!(BN_lshift(sdiv,divisor,norm_shift)))
{
goto err;
}
sdiv->neg=0;
norm_shift+=BN_BITS2;
if (!(BN_lshift(snum,num,norm_shift)))
{
goto err;
}
snum->neg=0;
div_n=sdiv->top;
num_n=snum->top;
loop=num_n-div_n;
/* Lets setup a 'window' into snum
* This is the part that corresponds to the current
* 'area' being divided */
BN_init(&wnum);
wnum.d= &(snum->d[loop]);
wnum.top= div_n;
wnum.dmax= snum->dmax+1; /* a bit of a lie */
/* Get the top 2 words of sdiv */
/* i=sdiv->top; */
d0=sdiv->d[div_n-1];
d1=(div_n == 1)?0:sdiv->d[div_n-2];
/* pointer to the 'top' of snum */
wnump= &(snum->d[num_n-1]);
/* Setup to 'res' */
res->neg= (num->neg^divisor->neg);
if (!bn_wexpand(res,(loop+1)))
{
goto err;
}
res->top=loop;
resp= &(res->d[loop-1]);
/* space for temp */
if (!bn_wexpand(tmp,(div_n+1)))
{
goto err;
}
if (BN_ucmp(&wnum,sdiv) >= 0)
{
if (!BN_usub(&wnum,&wnum,sdiv))
{
goto err;
}
*resp=1;
res->d[res->top-1]=1;
}
else
{
res->top--;
}
if (res->top == 0)
{
res->neg = 0;
}
resp--;
for (i=0; i<loop-1; i++)
{
BN_ULONG q,l0;
BN_ULONG n0,n1,rem=0;
n0=wnump[0];
n1=wnump[-1];
if (n0 == d0)
{
q=BN_MASK2;
}
else /* n0 < d0 */
{
BN_ULLONG t2;
q=(BN_ULONG)(((((BN_ULLONG)n0)<<BN_BITS2)|n1)/d0);
/*
* rem doesn't have to be BN_ULLONG. The least we
* know it's less that d0, isn't it?
*/
rem=(n1-q*d0)&BN_MASK2;
t2=(BN_ULLONG)d1*q;
for (;;)
{
if (t2 <= ((((BN_ULLONG)rem)<<BN_BITS2)|wnump[-2]))
{
break;
}
q--;
rem += d0;
if (rem < d0)
{
break; /* don't let rem overflow */
}
t2 -= d1;
}
}
l0=bn_mul_words(tmp->d,sdiv->d,div_n,q);
wnum.d--; wnum.top++;
tmp->d[div_n]=l0;
for (j=div_n+1; j>0; j--)
{
if (tmp->d[j-1])
{
break;
}
}
tmp->top=j;
j=wnum.top;
if (!BN_sub(&wnum,&wnum,tmp))
{
goto err;
}
snum->top=snum->top+wnum.top-j;
if (wnum.neg)
{
q--;
j=wnum.top;
if (!BN_add(&wnum,&wnum,sdiv))
{
goto err;
}
snum->top+=wnum.top-j;
}
*(resp--)=q;
wnump--;
}
if (rm != NULL)
{
/* Keep a copy of the neg flag in num because if rm==num
* BN_rshift() will overwrite it.
*/
int neg = num->neg;
BN_rshift(rm,snum,norm_shift);
if (!BN_is_zero(rm))
{
rm->neg = neg;
}
}
BN_CTX_end(ctx);
return(1);
err:
BN_CTX_end(ctx);
return(0);
}
#endif
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