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C ECC 双相加

转载 作者:行者123 更新时间:2023-11-30 19:09:07 26 4
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我目前正在使用双相加算法编写椭圆曲线加密的 C 代码。我面临着我不明白的段错误问题。我希望你们中有人能有一个想法。

#include "lib/include/gmp.h"
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <openssl/sha.h>

typedef struct{
mpz_t p;
mpz_t a;
mpz_t b;
mpz_t gx;
mpz_t gy;
} ECC;

static mpz_t *inverse_y1, *inverse_y2, *tx, *ty;
ECC secp256k1;

/*
* Initializes the secp256k1 curve.
*/
void initSECP256K1(){
mpz_set_str(secp256k1.p, "FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F", 16);
mpz_init(secp256k1.a);
mpz_set_str(secp256k1.b, "7", 10);
mpz_set_str(secp256k1.gx, "79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798", 16);
mpz_set_str(secp256k1.gy, "483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8", 16);
inverse_y1 = malloc(sizeof(mpz_t));
inverse_y2 = malloc(sizeof(mpz_t));
mpz_set_str(*inverse_y1, "483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8", 16);
mpz_set_str(*inverse_y2, "b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777", 16);
}

/*
* Calculates the s value for ecc point and doubling.
* s = (y2 - y1) / (x2 - x1) mod p if P != Q
* s = (3 * x1 ^ 2 + a) / (2 * y1) mod p if P = Q
* 5 parameters. First one is mpz_t pointer to save the result.
* The other 4 are of type pointer, representing the coordinates.
*/
void calculate_s(mpz_t *s, mpz_t *x1, mpz_t *y1, mpz_t *x2, mpz_t *y2){
mpz_t *s1 = malloc(sizeof(mpz_t));
mpz_init(*s1);
//Point Addition
if(mpz_cmp(*x1, *x2) != 0 || mpz_cmp(*y1, *y2) != 0){
mpz_sub(*s, *y2, *y1);
mpz_sub(*s1, *x2, *x1);
}else{
//Point doubling
mpz_powm_ui(*s, *x1, 2, secp256k1.p);
mpz_mul_si(*s, *s, 3);
mpz_mul_si(*s1, *y1, 2);
}
mpz_invert(*s1, *s1, secp256k1.p);
mpz_mul(*s, *s, *s1);
mpz_clear(*s1);
}



/*
* ECC - Point add and Point double
* It takes 6 parameters, each of type mpz_t pointer.
* The result is stored in the first two pointers.
* The third and forth pointer represent the first coordinate.
* The last two pointer are the second coordinate.
* It calculates the new coordinates as follows:
* x3 = s ^ 2 - x1 - x2 mod p
* y3 = s * (x1 - x3) - y1 mod p
*/
int ecc_papd(mpz_t *x3, mpz_t *y3, mpz_t *x1, mpz_t *y1, mpz_t *x2, mpz_t *y2){
mpz_t *s = malloc(sizeof(mpz_t));
mpz_init(*s);
//calculates s value
calculate_s(s, x1, y1, x2, y2);
//x3 = s^2 - x1 - x2 mpd p
mpz_powm_ui(*x3, *s, 2, secp256k1.p);
mpz_sub(*x3, *x3, *x1);
mpz_sub(*x3, *x3, *x2);
mpz_mod(*x3, *x3, secp256k1.p);
//y3 = s * (x1 - x3) - y1 mod p
mpz_sub(*y3, *x1, *x3);
mpz_mul(*y3, *y3, *s);
mpz_sub(*y3, *y3, *y1);
mpz_mod(*y3, *y3, secp256k1.p);
mpz_clear(*s);
//check if the result is the infinity point
if(mpz_cmp(*x1, *x2) == 0){
if((mpz_cmp(*y1, *inverse_y1) == 0 && mpz_cmp(*y2, *inverse_y2) == 0) ||
(mpz_cmp(*y2, *inverse_y1) == 0 && mpz_cmp(*y1, *inverse_y2) == 0)){
//printf("INFINITY");
return 1;
}
}
return 0;
}


void ecc_double_add(mpz_t *rx, mpz_t *ry, mpz_t *x, mpz_t *y, mpz_t d){
mpz_t *tx = malloc(sizeof(mpz_t));
mpz_t *ty = malloc(sizeof(mpz_t));
mpz_set(*tx, *x);
mpz_set(*ty, *y);
//returns the amount of bits the number has
//sub 2 because it starts to count from 1 and we remove the msb, too.
int bits = (int) mpz_sizeinbase(d, 2) - 2;
int bit, infinity = 0;
//check if bits is -1, (case d=1). If yes, set the base point as solution
if(bits < 0){
mpz_set(*rx, *x);
mpz_set(*ry, *y);
}
for(; bits >= 0; bits--){
bit = mpz_tstbit(d, bits);
if(infinity == 0){
infinity = ecc_papd(rx, ry, tx, ty, tx, ty);
mpz_set(*tx, *rx);
mpz_set(*ty, *ry);
}
//point addition
if(bit == 1){
if(infinity == 0){
infinity = ecc_papd(rx, ry, tx, ty, x, y);
mpz_set(*tx, *rx);
mpz_set(*ty, *ry);
}else{
mpz_set(*tx, *x);
mpz_set(*ty, *y);
infinity = 0;
}
}
}
free(tx);
free(ty);
}

int main(){
initSECP256K1();
mpz_t *rx = malloc(sizeof(mpz_t));
mpz_t *ry = malloc(sizeof(mpz_t));
//tx = malloc(sizeof(mpz_t));
//ty = malloc(sizeof(mpz_t));
mpz_t d;
mpz_init(*rx);
mpz_init(*ry);
mpz_init(d);

int i = 0;
for(; i < 2; i++){
gmp_printf ("d %Zd\n", d);
ecc_double_add(rx, ry, &secp256k1.gx, &secp256k1.gy, d);
gmp_printf ("R.x %Zx\n", rx);
gmp_printf ("R.y %Zx\n", ry);
printf("\n");
mpz_add_ui(d, d, 1);
}
//mpz_clear(*tx);
//mpz_clear(*ty);
mpz_clear(*rx);
mpz_clear(*ry);

return 0;
}

问题出在 ecc_double_add 函数内,使用 mpz_t *tx 和 mpz_t *ty 变量。当我运行代码时,出现段错误。因此,我在函数外部初始化了变量(在主函数中注释掉)。一旦我这样做了,它就会起作用。我很困惑,不知道为什么会收到段错误。我的 C 语言知识不太好,我正在努力提高。我感谢任何帮助:)

*编辑:我正在使用 GNU MP 库来完成此任务,它位于:https://gmplib.org/

最佳答案

我一直在修补,发现 gmp_printf 函数导致了问题。一旦我将其从 for 循环中删除,代码就可以工作了。我用不带参数的 printf 和 gmp_printf 替换了它,代码开始工作。我不确定为什么 gmp_printf 会出现问题,但我发现了这个: https://gmplib.org/list-archives/gmp-bugs/2011-July/002304.html

我用

替换了 gmp_printf
printf("Result: %s\n", mpz_get_str(NULL, 16, *rx));

这不是一个很好的解决方案,但对于测试来说应该足够了。

关于C ECC 双相加,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43748136/

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