Question

我有问题，希望您能帮助我。

我有一个用C编写的函数，该函数返回哈希值。我的令人头疼的是，当我从另一个工具执行程序时，要花很多时间才能运行，这可能是因为在函数内部运行了一个在SHA256中哈希值的命令，所以我想知道是否还有另一种方法可以执行此操作，也许是函数之类的东西。

这是我所拥有的：

const char *EncryptSHA256 (char *Arg1) {  
    char command[128];
    char result[512];

    //I want to replace from here
    snprintf(command, sizeof command, "echo -n %s | sha256sum | cut -c1-64",Arg1);  

    FILE *fpipe;

    if (0 == (fpipe = (FILE*)popen(command, "r"))) {
        perror("popen() failed.");
        exit(1);
    }

    fread(result, 1, 512, fpipe);

    pclose(fpipe);
    const char *sha256 = &result[0];
    //to here

    return sha256;
}

Answer 1

您的代码具有未定义的行为，因为您返回了指向result（具有自动存储功能的本地数组）的指针。调用者从此数组读取具有未定义的行为。

您至少应将result设为静态，以便在EncryptSHA256返回其调用方之后，其内容仍然可读。

关于方法的低效率，这是SHA256的公共领域实现，您可以在程序内部直接使用它：

/* public domain sha256 implementation based on fips180-3 */

#include <stddef.h>
#include <stdint.h>
#include <string.h>

/* Public API */

struct sha256 {
    uint64_t len;    /* processed message length */
    uint32_t h[8];   /* hash state */
    uint8_t buf[64]; /* message block buffer */
};

/* reset state */
void sha256_init(struct sha256 *s);
/* process message */
void sha256_update(struct sha256 *s, const void *m, size_t len);
/* get message digest */
/* state is ruined after sum, keep a copy if multiple sum is needed */
/* part of the message might be left in s, zero it if secrecy is needed */
void sha256_sum(struct sha256 *s, uint8_t md[32]);

/* Implementation */

static uint32_t ror(uint32_t n, int k) {
    return (n >> k) | (n << (32 - k));
}
#define Ch(x,y,z)  (z ^ (x & (y ^ z)))
#define Maj(x,y,z) ((x & y) | (z & (x | y)))
#define S0(x)      (ror(x,2) ^ ror(x,13) ^ ror(x,22))
#define S1(x)      (ror(x,6) ^ ror(x,11) ^ ror(x,25))
#define R0(x)      (ror(x,7) ^ ror(x,18) ^ (x>>3))
#define R1(x)      (ror(x,17) ^ ror(x,19) ^ (x>>10))

static const uint32_t K[64] = {
    0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
    0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
    0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
    0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
    0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
    0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
    0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
    0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
    0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
    0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
    0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
    0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
    0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
    0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
    0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
    0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2
};

static void processblock(struct sha256 *s, const uint8_t *buf) {
    uint32_t W[64], t1, t2, a, b, c, d, e, f, g, h;
    int i;

    for (i = 0; i < 16; i++) {
        W[i]  = (uint32_t)buf[4 * i + 0] << 24;
        W[i] |= (uint32_t)buf[4 * i + 1] << 16;
        W[i] |= (uint32_t)buf[4 * i + 2] << 8;
        W[i] |= buf[4 * i + 3];
    }
    for (; i < 64; i++)
        W[i] = R1(W[i-2]) + W[i-7] + R0(W[i-15]) + W[i-16];
    a = s->h[0];
    b = s->h[1];
    c = s->h[2];
    d = s->h[3];
    e = s->h[4];
    f = s->h[5];
    g = s->h[6];
    h = s->h[7];
#define ROUND(a,b,c,d,e,f,g,h,i) \
        t1 = h + S1(e) + Ch(e,f,g) + K[i] + W[i]; \
        t2 = S0(a) + Maj(a,b,c); \
        d += t1; \
        h = t1 + t2;
    for (i = 0; i < 64; ) {
        ROUND(a, b, c, d, e, f, g, h, i); i++;
        ROUND(h, a, b, c, d, e, f, g, i); i++;
        ROUND(g, h, a, b, c, d, e, f, i); i++;
        ROUND(f, g, h, a, b, c, d, e, i); i++;
        ROUND(e, f, g, h, a, b, c, d, i); i++;
        ROUND(d, e, f, g, h, a, b, c, i); i++;
        ROUND(c, d, e, f, g, h, a, b, i); i++;
        ROUND(b, c, d, e, f, g, h, a, i); i++;
    }
#undef ROUND
    s->h[0] += a;
    s->h[1] += b;
    s->h[2] += c;
    s->h[3] += d;
    s->h[4] += e;
    s->h[5] += f;
    s->h[6] += g;
    s->h[7] += h;
}

static void pad(struct sha256 *s) {
    unsigned r = s->len % 64;

    s->buf[r++] = 0x80;
    if (r > 56) {
        memset(s->buf + r, 0, 64 - r);
        r = 0;
        processblock(s, s->buf);
    }
    memset(s->buf + r, 0, 56 - r);
    s->len *= 8;
    s->buf[56] = s->len >> 56;
    s->buf[57] = s->len >> 48;
    s->buf[58] = s->len >> 40;
    s->buf[59] = s->len >> 32;
    s->buf[60] = s->len >> 24;
    s->buf[61] = s->len >> 16;
    s->buf[62] = s->len >> 8;
    s->buf[63] = s->len;
    processblock(s, s->buf);
}

void sha256_init(struct sha256 *s) {
    s->len = 0;
    s->h[0] = 0x6a09e667;
    s->h[1] = 0xbb67ae85;
    s->h[2] = 0x3c6ef372;
    s->h[3] = 0xa54ff53a;
    s->h[4] = 0x510e527f;
    s->h[5] = 0x9b05688c;
    s->h[6] = 0x1f83d9ab;
    s->h[7] = 0x5be0cd19;
}

void sha256_sum(struct sha256 *s, uint8_t md[20]) {
    int i;

    pad(s);
    for (i = 0; i < 8; i++) {
        md[4 * i + 0] = s->h[i] >> 24;
        md[4 * i + 1] = s->h[i] >> 16;
        md[4 * i + 2] = s->h[i] >> 8;
        md[4 * i + 3] = s->h[i];
    }
}

void sha256_update(struct sha256 *s, const void *m, unsigned long len) {
    const uint8_t *p = m;
    unsigned r = s->len % 64;

    s->len += len;
    if (r) {
        if (len < 64 - r) {
            memcpy(s->buf + r, p, len);
            return;
        }
        memcpy(s->buf + r, p, 64 - r);
        len -= 64 - r;
        p += 64 - r;
        processblock(s, s->buf);
    }
    for (; len >= 64; len -= 64, p += 64)
        processblock(s, p);
    memcpy(s->buf, p, len);
}

您将功能更改为此：

const char *EncryptSHA256(char *Arg1) {  
    struct sha256 s;
    unsigned char md[32];
    static char result[65];

    sha256_init(&s);
    sha256_update(&s, Arg1, strlen(Arg1));
    sha256_sum(&s, md);
    for (int i = 0; i < 32; i++) {
        sprintf(result + i * 2, "%02x", md[i]);
    }
    return result;
}

如果更方便，您也可以更改API以传递32个无符号字符的数组以获取二进制形式。

来自C函数的哈希值

1 个答案: