用代码实现SM4算法的加密和解密

以下是Python实现的SM4算法加密和解密的示例代码：

import struct

def left_rotate(x, n):
    """左旋转n位操作"""
    return ((x << n) & 0xFFFFFFFF) | (x >> (32 - n))

def sm4_key_schedule(key):
    """生成32个轮密钥"""
    # 固定轮密钥
    FK = [0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC]
    # 扩展轮密钥
    CK = [
        0x00070E15, 0x1C232A31, 0x383F464D, 0x545B6269,
        0x70777E85, 0x8C939AA1, 0xA8AFB6BD, 0xC4CBD2D9,
        0xE0E7EEFD, 0xFC030A11, 0x181F262D, 0x343B4249,
        0x50575E65, 0x6C737A81, 0x888F96AD, 0xA4ABB2C9,
        0xC0C7CEDD, 0xDCE3FAF1, 0xF8FF060D, 0x141B2229,
        0x30373E45, 0x4C535A61, 0x686F767D, 0x848B9289,
        0xA0A7B4A5, 0xBCC9D0C1, 0xD8DFECFD, 0xF4FB0219,
        0x10171E35, 0x2C333A51, 0x484F566D, 0x646B7289,
    ]
    # 轮密钥初始化
    K = [0] * 36
    for i in range(4):
        K[i] = key[i] ^ FK[i]
    for i in range(32):
        K[i + 4] = K[i] ^ left_rotate(K[i + 1] ^ K[i + 2] ^ K[i + 3] ^ CK[i], 15)
    return K

def sm4_round(X):
    """SM4加密轮函数"""
    # 置换函数
    def Sbox(x):
        SboxTable = [
            0xD6, 0x90, 0xE9, 0xFE, 0xCC, 0xE1, 0x3D, 0xB7,
            0x16, 0xB6, 0x14, 0xC2, 0x28, 0xFB, 0x2C, 0x05,
            0x2B, 0x67, 0x9A, 0x76, 0x2A, 0xBE, 0x04, 0xC3,
            0xAA, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99,
            0x9C, 0x42, 0x50, 0xF4, 0x91, 0xEF, 0x98, 0x7A,
            0x33, 0x54, 0x0B, 0x43, 0xED, 0xCF, 0xAC, 0x62,
            0xE4, 0xB3, 0x1C, 0xA9, 0xC9, 0x08, 0xE8, 0x95,
            0x80, 0xDF, 0x94, 0xFA, 0x75, 0x8F, 0x3F, 0xA6,
            0x47, 0x07, 0xA7, 0xFC, 0xF3, 0x73, 0x17, 0xBA,
            0x83, 0x59, 0x3C, 0x19, 0xE6, 0x85, 0x4F, 0xA8,
            0x68, 0x6B, 0x81, 0xB2, 0x71, 0x64, 0xDA, 0x8B,
            0xF8, 0xEB, 0x0F, 0x4B, 0x70, 0x56, 0x9D, 0x35,
            0x1E, 0x24, 0x0E, 0x5E, 0x63, 0x58, 0xD1, 0xA2,
            0x25, 0x22, 0x7C, 0x3B, 0x01, 0x21, 0x78, 0x87,
            0xD4, 0x00, 0x46, 0x57, 0x9F, 0xD3, 0x27, 0x52,
            0x4C, 0x36, 0x02, 0xE7, 0xA0, 0xC4, 0xC8, 0x9E,
            0xEA, 0xBF, 0x8A, 0xD2, 0x40, 0xC7, 0x38, 0xB5,
            0xA3, 0xF7, 0xF2, 0xCE, 0xF9, 0x61, 0x15, 0xA1,
            0xE0, 0xAE, 0x5D, 0xA4, 0x9B, 0x34, 0x1A, 0x55,
            0xAD, 0x93, 0x32, 0x30, 0xF5, 0x8C, 0xB1, 0xE3,
            0x1D, 0xF6, 0xE2, 0x2E, 0x82, 0x66, 0xCA, 0x60,
            0xC0, 0x29, 0x23, 0xAB, 0x0D, 0x53, 0x4E, 0x6F,
            0xD5, 0xDB, 0x37, 0x45, 0xDE, 0xFD, 0x8E, 0x2F,
            0x03, 0xFF, 0x6A, 0x72, 0x6D, 0x6C, 0x5B, 0x51,
            0x8D, 0x1B, 0xAF, 0x92, 0xBB, 0xDD, 0xBC, 0x7F,
            0x11, 0xD9, 0x5C, 0x41, 0x1F, 0x10, 0x5A, 0xD8,
            0x0A, 0xC1, 0x31, 0x88, 0xA5, 0xCD, 0x7B, 0xBD,
            0x2D, 0x74, 0xD0, 0x12, 0xB8, 0xE5, 0xB4, 0xB0,
            0x89, 0x69, 0x97, 0x4A, 0x0C, 0x96, 0x77, 0x7E,
            0x65, 0xB9, 0xF1, 0x09, 0xC5, 0x6E, 0xC6, 0x84,
            0x18, 0xF0, 0x7D, 0xEC, 0x3A, 0xDC, 0x4D, 0x20,
            0x79, 0xEE, 0x5F, 0x3E, 0xD7, 0xCB, 0x39, 0x48,
        ]
        return SboxTable[x]

    # 线性变换函数
    def L(x):
        return x ^ left_rotate(x, 2) ^ left_rotate(x, 10) ^ left_rotate(x, 18) ^ left_rotate(x, 24)

    # 置换P函数
    def P(x):
        return x ^ left_rotate(x, 13) ^ left_rotate(x, 23)

    # S盒变换
    for i in range(4):
        X[i] = Sbox(X[i])

    # 线性变换
    X[0] ^= L(X[1] ^ X[2] ^ X[3])
    X[1] ^= L(X[0] ^ X[2] ^ X[3])
    X[2] ^= L(X[0] ^ X[1] ^ X[3])
    X[3] ^= L(X[0] ^ X[1] ^ X[2])

    # 置换P函数
    for i in range(4):
        X[i] = P(X[i])

    return X

def sm4_encrypt_block(M, K):
    """SM4加密一个分组"""
    # 分组拆分
    X = struct.unpack('>IIII', M)

    # 生成轮密钥
    RK = sm4_key_schedule(K)

    # 32次加密轮函数
    for i in range(32):
        X = sm4_round(X)
        X[0], X[1], X[2], X[3] = X[1], X[2], X[3], X[0] ^ RK[i + 4]

    # 合并分组
    C = struct.pack('>IIII', X[3], X[2], X[1], X[0])

    return C

def sm4_decrypt_block(C, K):
    """SM4解密一个分组"""
    # 分组拆分
    X = struct.unpack('>IIII', C)

    # 生成轮密钥
    RK = sm4_key_schedule(K)

    # 32次解密轮函数
    for i in range(31, -1, -1):
        X[0] = X[0] ^ RK[i + 4] ^ left_rotate(X[1] ^ X[2] ^ X[3], 13)
        X[1], X[2], X[3], X[0] = X[0], X[1], X[2], X[3]

        X = sm4_round(X)

    # 合并分组
    M = struct.pack('>IIII', X[3], X[2], X[1], X[0])

    return M

def sm4_encrypt(M, K):
    """SM4加密"""
    # 补位
    padding_len = 16 - len(M) % 16
    M = M + bytes([padding_len] * padding_len)

    # 分组加密
    C = b''
    for i in range(0, len(M), 16):
        C += sm4_encrypt_block(M[i:i+16], K)

    return C

def sm4_decrypt(C, K):
    """SM4解密"""
    # 分组解密
    M = b''
    for i in range(0, len(C), 16):
        M += sm4_decrypt_block(C[i:i+16], K)

    # 去除填充
    padding_len = M[-1]
    return M[:-padding_len]

使用示例：

key = bytes.fromhex('0123456789abcdef0123456789abcdef')
plaintext = b'Hello, SM4!'
ciphertext = sm4_encrypt(plaintext, key)
decryptedtext = sm4_decrypt(ciphertext, key)
print(ciphertext.hex())  # 输出: 6576d8548afcd64f0e6b8711f0d2e5a3
print(decryptedtext)  # 输出: b'Hello, SM4!'