来自C函数的哈希值

Question

我有一个问题，希望你能帮助我。

我有一个用C编写的函数，它返回一个哈希值。我的头痛是当我从另一个工具执行程序时需要花费大量的时间来运行，可能是因为在我的函数中我运行了一个命令，它在SHA256中散列我的值，所以我想知道是否有其他方法可以执行它，也许是一个功能或类似的东西。

这是我有的：

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，因为您返回指向result的指针，result是具有自动存储的本地数组。调用者从此数组中读取具有未定义的行为。

你应该至少使EncryptSHA256静态，以便在

/* 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);
}

返回其调用者之后其内容仍然可读。

关于方法的低效率，这里是SHA256的公共域实现，您可以直接在程序中使用：

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;
}

您可以将功能更改为：

qazxswpoi

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