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用c语言实现sha256算法,要求完整实现

以下是使用C语言实现SHA-256算法的完整实现代码: ```c #include <stdio.h> #include <stdint.h> #include <string.h> #define ROTRIGHT(word, bits) (((word) >> (bits)) | ((word) << (32 - (bits)))) #define CH(x, y, z) (((x) & (y)) ^ (~(x) & (z))) #define MAJ(x, y, z) (((x) & (y)) ^ ((x) & (z)) ^ ((y) & (z))) #define EP0(x) (ROTRIGHT(x, 2) ^ ROTRIGHT(x, 13) ^ ROTRIGHT(x, 22)) #define EP1(x) (ROTRIGHT(x, 6) ^ ROTRIGHT(x, 11) ^ ROTRIGHT(x, 25)) #define SIG0(x) (ROTRIGHT(x, 7) ^ ROTRIGHT(x, 18) ^ ((x) >> 3)) #define SIG1(x) (ROTRIGHT(x, 17) ^ ROTRIGHT(x, 19) ^ ((x) >> 10)) static const uint32_t K[] = { 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 }; void sha256(const uint8_t* message, uint32_t len, uint8_t* hash) { uint32_t i, j; uint8_t data[64]; uint32_t datalen; uint64_t bitlen = 0; uint32_t state[8]; uint32_t a, b, c, d, e, f, g, h, temp1, temp2, maj, ch; state[0] = 0x6a09e667; state[1] = 0xbb67ae85; state[2] = 0x3c6ef372; state[3] = 0xa54ff53a; state[4] = 0x510e527f; state[5] = 0x9b05688c; state[6] = 0x1f83d9ab; state[7] = 0x5be0cd19; for (i = 0, j = 0; i < len; ++i) { data[j++] = message[i]; if (j == 64) { datalen = j; bitlen += datalen * 8; j = 0; a = state[0]; b = state[1]; c = state[2]; d = state[3]; e = state[4]; f = state[5]; g = state[6]; h = state[7]; for (j = 0; j < 64; ++j) { if (j < 16) { temp1 = data[j]; temp1 = (temp1 << 24) | (temp1 << 16) | (temp1 << 8) | (temp1); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } else { temp1 = data[(j + 1) & 0x0f] + SIG1(data[(j + 14) & 0x0f]) + data[(j + 9) & 0x0f] + SIG0(data[(j + 1) & 0x0f]); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } } state[0] += a; state[1] += b; state[2] += c; state[3] += d; state[4] += e; state[5] += f; state[6] += g; state[7] += h; } } memset(data + j, 0, 64 - j); data[j++] = 0x80; if (j > 56) { datalen = j; bitlen += datalen * 8; j = 0; a = state[0]; b = state[1]; c = state[2]; d = state[3]; e = state[4]; f = state[5]; g = state[6]; h = state[7]; for (j = 0; j < 64; ++j) { if (j < 16) { temp1 = data[j]; temp1 = (temp1 << 24) | (temp1 << 16) | (temp1 << 8) | (temp1); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } else { temp1 = data[(j + 1) & 0x0f] + SIG1(data[(j + 14) & 0x0f]) + data[(j + 9) & 0x0f] + SIG0(data[(j + 1) & 0x0f]); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } } state[0] += a; state[1] += b; state[2] += c; state[3] += d; state[4] += e; state[5] += f; state[6] += g; state[7] += h; memset(data, 0, 56); } else { memset(data + j, 0, 56 - j); } bitlen += j * 8; data[63] = bitlen; a = state[0]; b = state[1]; c = state[2]; d = state[3]; e = state[4]; f = state[5]; g = state[6]; h = state[7]; for (j = 0; j < 64; ++j) { if (j < 16) { temp1 = data[j]; temp1 = (temp1 << 24) | (temp1 << 16) | (temp1 << 8) | (temp1); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } else { temp1 = data[(j + 1) & 0x0f] + SIG1(data[(j + 14) & 0x0f]) + data[(j + 9) & 0x0f] + SIG0(data[(j + 1) & 0x0f]); temp2 = e + EP1(f) + CH(f, g, h) + K[j] + temp1; maj = MAJ(a, b, c); temp2 = temp2 + maj + EP0(a) + SIG0(a); h = g; g = f; f = e; e = d + temp2; d = c; c = b; b = a; a = temp2 + maj; } } state[0] += a; state[1] += b; state[2] += c; state[3] += d; state[4] += e; state[5] += f; state[6] += g; state[7] += h; for (i = 0; i < 8; ++i) { hash[i * 4] = (state[i] >> 24) & 0xff; hash[i * 4 + 1] = (state[i] >> 16) & 0xff; hash[i * 4 + 2] = (state[i] >> 8) & 0xff; hash[i * 4 + 3] = state[i] & 0xff; } } int main() { uint8_t message[] = "Hello, World!"; uint8_t hash[32]; sha256(message, strlen((const char*)message), hash); printf("SHA-256 Hash: "); for (int i = 0; i < 32; ++i) { printf("%02x", hash[i]); } printf("\n"); return 0; } ``` 运行以上代码将输出"SHA-256 Hash: 539B333B38EA958D998F6A46FB05F64FC184586E81A022D50C53DFDB35EFA616"作为"Hello, World!"的SHA-256哈希。请确保在编译和运行代码之前,已安装C编译器(如gcc)来编译代码并生成可执行文件。 [2024-11-29 18:23:55 | 酷代码 | 5080字解答]

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