用c语言实现sha256算法，要求包含init update finish三个过程，输入输出都是uint8_t类型的数组

实现SHA-256算法需要一些基本的位操作和常量定义。以下是一个简单的C语言实现，包含`init`、`update`和`finish`三个过程，输入输出都是`uint8_t`类型的数组。 ```c #include <stdint.h> #include <string.h> // SHA-256 constants 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 }; typedef struct { uint8_t data[64]; uint32_t datalen; unsigned long long bitlen; uint32_t state[8]; } SHA256_CTX; void sha256_transform(SHA256_CTX *ctx, const uint8_t data[]); void sha256_init(SHA256_CTX *ctx); void sha256_update(SHA256_CTX *ctx, const uint8_t data[], size_t len); void sha256_final(SHA256_CTX *ctx, uint8_t hash[]); void sha256_transform(SHA256_CTX *ctx, const uint8_t data[]) { uint32_t a, b, c, d, e, f, g, h, i, j, t1, t2, m[64]; for (i = 0, j = 0; i < 16; ++i, j += 4) m[i] = (data[j] << 24) | (data[j + 1] << 16) | (data[j + 2] << 8) | (data[j + 3]); for ( ; i < 64; ++i) m[i] = m[i - 16] + ((m[i - 15] >> 7 | m[i - 15] << (32 - 7)) ^ (m[i - 15] >> 18 | m[i - 15] << (32 - 18)) ^ (m[i - 15] >> 3)) + m[i - 7] + ((m[i - 2] >> 17 | m[i - 2] << (32 - 17)) ^ (m[i - 2] >> 19 | m[i - 2] << (32 - 19)) ^ (m[i - 2] >> 10)); a = ctx->state[0]; b = ctx->state[1]; c = ctx->state[2]; d = ctx->state[3]; e = ctx->state[4]; f = ctx->state[5]; g = ctx->state[6]; h = ctx->state[7]; for (i = 0; i < 64; ++i) { t1 = h + ((e >> 6 | e << (32 - 6)) ^ (e >> 11 | e << (32 - 11)) ^ (e >> 25 | e << (32 - 25))) + ((e & f) ^ (~e & g)) + k[i] + m[i]; t2 = ((a >> 2 | a << (32 - 2)) ^ (a >> 13 | a << (32 - 13)) ^ (a >> 22 | a << (32 - 22))) + ((a & b) ^ (a & c) ^ (b & c)); h = g; g = f; f = e; e = d + t1; d = c; c = b; b = a; a = t1 + t2; } ctx->state[0] += a; ctx->state[1] += b; ctx->state[2] += c; ctx->state[3] += d; ctx->state[4] += e; ctx->state[5] += f; ctx->state[6] += g; ctx->state[7] += h; } void sha256_init(SHA256_CTX *ctx) { ctx->datalen = 0; ctx->bitlen = 0; ctx->state[0] = 0x6a09e667; ctx->state[1] = 0xbb67ae85; ctx->state[2] = 0x3c6ef372; ctx->state[3] = 0xa54ff53a; ctx->state[4] = 0x510e527f; ctx->state[5] = 0x9b05688c; ctx->state[6] = 0x1f83d9ab; ctx->state[7] = 0x5be0cd19; } void sha256_update(SHA256_CTX *ctx, const uint8_t data[], size_t len) { for (size_t i = 0; i < len; ++i) { ctx->data[ctx->datalen] = data[i]; ctx->datalen++; if (ctx->datalen == 64) { [2024-12-01 09:27:53 | AI写代码神器 | 2048点数解答]