我提供此示例应用程序来显示我的问题
#include <stdio.h>
#include <stdlib.h>
#include <openssl/ec.h>
#include <openssl/bn.h>
int main()
{
EC_KEY *pkey = NULL;
EC_POINT *pub_key = NULL;
const EC_GROUP *group = NULL;
BIGNUM start;
BIGNUM *res;
BN_CTX *ctx;
BN_init(&start);
ctx = BN_CTX_new();
res = &start;
BN_hex2bn(&res,"3D79F601620A6D05DB7FED883AB8BCD08A9101B166BC60166869DA5FC08D936E");
pkey = EC_KEY_new_by_curve_name(NID_secp256k1);
group = EC_KEY_get0_group(pkey);
pub_key = EC_POINT_new(group);
EC_KEY_set_private_key(pkey, res);
assert(EC_POINT_bn2point(group,res, pub_key, ctx)); // Null here
EC_KEY_set_public_key(pkey, pub_key);
return 0;
}
我要做的是从私钥显示公钥(应该是椭圆私钥)。 在遇到类似的问题之前,我不知道该怎么做
How do I feed OpenSSL random data for use in ECDSA signing?
我从哪里开始指出自己如何获取公钥并使用EC_POINT_bn2point而不是根据OpenSSL源在内部执行BN_hex2bn的hex2point。
那么,为什么EC_POINT_bn2point返回NULL?我正在认真考虑重新编译OpenSSL并设置一些调试例程来弄清楚它失败的原因。
答案 0 :(得分:4)
工作示例:
// using figures on: https://en.bitcoin.it/wiki/Technical_background_of_version_1_Bitcoin_addresses
// gcc -Wall ecdsapubkey.c -o ecdsapubkey -lcrypto
#include <stdio.h>
#include <stdlib.h>
#include <openssl/ec.h>
#include <openssl/obj_mac.h>
#include <openssl/bn.h>
int main()
{
EC_KEY *eckey = NULL;
EC_POINT *pub_key = NULL;
const EC_GROUP *group = NULL;
BIGNUM start;
BIGNUM *res;
BN_CTX *ctx;
BN_init(&start);
ctx = BN_CTX_new(); // ctx is an optional buffer to save time from allocating and deallocating memory whenever required
res = &start;
// BN_hex2bn(&res,"3D79F601620A6D05DB7FED883AB8BCD08A9101B166BC60166869DA5FC08D936E");
BN_hex2bn(&res,"18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725");
eckey = EC_KEY_new_by_curve_name(NID_secp256k1);
group = EC_KEY_get0_group(eckey);
pub_key = EC_POINT_new(group);
EC_KEY_set_private_key(eckey, res);
/* pub_key is a new uninitialized `EC_POINT*`. priv_key res is a `BIGNUM*`. */
if (!EC_POINT_mul(group, pub_key, res, NULL, NULL, ctx))
printf("Error at EC_POINT_mul.\n");
// assert(EC_POINT_bn2point(group, &res, pub_key, ctx)); // Null here
EC_KEY_set_public_key(eckey, pub_key);
char *cc = EC_POINT_point2hex(group, pub_key, 4, ctx);
char *c=cc;
int i;
for (i=0; i<130; i++) // 1 byte 0x42, 32 bytes for X coordinate, 32 bytes for Y coordinate
{
printf("%c", *c++);
}
printf("\n");
BN_CTX_free(ctx);
free(cc);
return 0;
}
另见http://wiki.openssl.org/index.php/Elliptic_Curve_Cryptography - 库