CBC填充预言攻击-CBC Padding Oracle Crypto Attack

CBC填充预言攻击需要一个能够提示解密结果填充正确与否的服务端（即填充预言机），以及完整的密文和IV（若无IV无法解密第一个分组密文，后续分组也可正常解密）

详见CAPEC-463: Padding Oracle Crypto Attack

https://capec.mitre.org/data/definitions/463.html

总的来说，攻击者可以通过不断尝试向填充预言机枚举密文，通过CBC加密和填充模式的结构，绕过密钥和解密机制猜测到密文解密后的中间值，与真实的C_{i-1}异或，即可得到明文。

接下来一步步详细讲解

首先列举一下我们已有的条件

密文：27a5d7ba48325193a6e27c33acf5ea14f4fc9d11c3dbde6afa6dc560e4a60ecbc2a58ef082b56464b4e172056795a87a

IV：000102030405060708090a0b0c0d0e0f
填充预言机：输入密文进行解密，可以通过状态码等方式明确解密后填充是否正常

国密标准SM4-CBC的填充方式是PKCS-7填充，即在末尾填充填充体的长度直到填满

CBC的解密机制，精确到每个分组，实际上是P_i = D_k(C_i) XOR C_{i-1}

假设现在我们攻击第一个密文分组来获取明文，因为有预言机存在，我们可以从0x00到0xFF枚举C_{i-1}[-1]，和真实的C_i拼接后放入预言机解密，直到提示填充正确。此时D_k(C_i)[-1] XOR C_{i-1}[-1]应当是0x01，然后我们再将0x01和猜测的C_{i-1}[-1]进行异或，即可得到真实的D_k(C_i)[-1]，然后再将D_k(C_i)[-1]和真实的IV[-1]进行异或，即可得到真实的P_i[-1]。

而且这个操作是可以递推的，我们可以控制猜测的C_{i-1}[-1]值，使得D_k(C_i)[-1] XOR C_{i-1}[-1]等于0x02后，枚举C_{i-1}[-2]的值，直到预言机再次给出正确回应，从而得到真实的P_i[-2]。以此类推，直到将整个分组全部解密。

Talk is cheap,上代码

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

SM4_SBOX = bytes([
    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,
])

# 系统参数 FK
SM4_FK = [0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC]

# 固定常数 CK
SM4_CK = [
    0x00070E15, 0x1C232A31, 0x383F464D, 0x545B6269,
    0x70777E85, 0x8C939AA1, 0xA8AFB6BD, 0xC4CBD2D9,
    0xE0E7EEF5, 0xFC030A11, 0x181F262D, 0x343B4249,
    0x50575E65, 0x6C737A81, 0x888F969D, 0xA4ABB2B9,
    0xC0C7CED5, 0xDCE3EAF1, 0xF8FF060D, 0x141B2229,
    0x30373E45, 0x4C535A61, 0x686F767D, 0x848B9299,
    0xA0A7AEB5, 0xBCC3CAD1, 0xD8DFE6ED, 0xF4FB0209,
    0x10171E25, 0x2C333A41, 0x484F565D, 0x646B7279,
]


def _sm4_tau(a):
    a = a & 0xFFFFFFFF
    return ((SM4_SBOX[(a >> 24) & 0xFF] << 24) |
            (SM4_SBOX[(a >> 16) & 0xFF] << 16) |
            (SM4_SBOX[(a >> 8) & 0xFF] << 8) |
            SM4_SBOX[a & 0xFF])


def _sm4_l(b):
    b = b & 0xFFFFFFFF
    return (b ^
            (((b << 2) | (b >> 30)) & 0xFFFFFFFF) ^
            (((b << 10) | (b >> 22)) & 0xFFFFFFFF) ^
            (((b << 18) | (b >> 14)) & 0xFFFFFFFF) ^
            (((b << 24) | (b >> 8)) & 0xFFFFFFFF)) & 0xFFFFFFFF


def _sm4_l_key(b):
    b = b & 0xFFFFFFFF
    return (b ^
            (((b << 13) | (b >> 19)) & 0xFFFFFFFF) ^
            (((b << 23) | (b >> 9)) & 0xFFFFFFFF)) & 0xFFFFFFFF


def _sm4_f(x0, x1, x2, x3, rk):
    return x0 ^ _sm4_l(_sm4_tau(x1 ^ x2 ^ x3 ^ rk))


def _sm4_key_expansion(key):
    k = []
    key_words = [int.from_bytes(key[i:i+4], 'big') for i in range(0, 16, 4)]
    for i in range(4):
        k.append(key_words[i] ^ SM4_FK[i])
    rk = []
    for i in range(32):
        rk.append(_sm4_l_key(_sm4_tau(k[i+1] ^ k[i+2] ^ k[i+3] ^ SM4_CK[i])))
        k.append((k[i] ^ rk[i]) & 0xFFFFFFFF)
    return rk


def _sm4_round(x, rk, encrypt=True):
    x = list(x)
    if not encrypt:
        rk = rk[::-1]  # 解密时轮密钥逆序
    for i in range(32):
        x.append(_sm4_f(x[i], x[i+1], x[i+2], x[i+3], rk[i]) & 0xFFFFFFFF)
    # 反序变换 R
    return [x[35] & 0xFFFFFFFF, x[34] & 0xFFFFFFFF,
            x[33] & 0xFFFFFFFF, x[32] & 0xFFFFFFFF]


def _bytes_to_words(data):
    return [int.from_bytes(data[i:i+4], 'big') for i in range(0, 16, 4)]


def _words_to_bytes(words):
    return b''.join((w & 0xFFFFFFFF).to_bytes(4, 'big') for w in words)


def sm4_encrypt_block(block, key):
    rk = _sm4_key_expansion(key)
    x = _bytes_to_words(block)
    result = _sm4_round(x, rk, encrypt=True)
    return _words_to_bytes(result)


def sm4_decrypt_block(block, key):
    rk = _sm4_key_expansion(key)
    x = _bytes_to_words(block)
    result = _sm4_round(x, rk, encrypt=False)
    return _words_to_bytes(result)


def pkcs7_pad(data, block_size=16):
    padding_length = block_size - (len(data) % block_size)
    padding = bytes([padding_length] * padding_length)
    return data + padding


def pkcs7_unpad(data):
    if not data:
        raise ValueError("Empty data")
    pad_len = data[-1]
    if pad_len == 0 or pad_len > 16:
        raise ValueError(f"Invalid padding length: {pad_len}")
    if data[-pad_len:] != bytes([pad_len] * pad_len):
        raise ValueError("Invalid padding bytes")
    return data[:-pad_len]


def sm4_cbc_encrypt(plaintext, key, iv):
    assert len(key) == 16 and len(iv) == 16
    padded = pkcs7_pad(plaintext)
    ciphertext = b''
    prev = iv
    for i in range(0, len(padded), 16):
        block = padded[i:i+16]
        xored = bytes(a ^ b for a, b in zip(block, prev))
        enc = sm4_encrypt_block(xored, key)
        ciphertext += enc
        prev = enc
    return ciphertext


def sm4_cbc_decrypt(ciphertext, key, iv):
    assert len(key) == 16 and len(iv) == 16
    assert len(ciphertext) % 16 == 0
    plaintext = b''
    prev = iv
    for i in range(0, len(ciphertext), 16):
        block = ciphertext[i:i+16]
        dec = sm4_decrypt_block(block, key)
        xored = bytes(a ^ b for a, b in zip(dec, prev))
        plaintext += xored
        prev = block
    return plaintext

class PaddingOracle:
    def __init__(self, key, iv):
        self.key = key
        self.iv = iv

    def check(self, ciphertext):
        try:
            decrypted = sm4_cbc_decrypt(ciphertext, self.key, self.iv)
            pkcs7_unpad(decrypted)
            return True
        except Exception:
            return False

class PaddingOracleAttacker:

    def __init__(self, oracle):
        self.oracle = oracle
        self.block_size = 16

    def attack_block(self, prev_block, target_block, verbose=True):
        if verbose:
            print(f"\n{'='*60}")
            print(f"开始攻击密文块: {target_block.hex()}")
            print(f"前驱块:        {prev_block.hex()}")

        # Decrypt(C_i) 的中间值（解密后的值，未与前一个块异或）
        intermediate = bytearray(self.block_size)
        # 恢复的明文块
        plaintext = bytearray(self.block_size)

        # 从右到左逐字节攻击
        for byte_pos in range(self.block_size - 1, -1, -1):
            padding_value = self.block_size - byte_pos  # 当前期望的填充值

            if verbose:
                print(f"\n  [攻击第 {byte_pos:2d} 字节] 期望填充: 0x{padding_value:02x}")

            # 构造伪造的前驱块
            fake_prev = bytearray(self.block_size)

            # 对于已经破解的字节（右侧），设置使得它们产生正确的填充
            for p in range(byte_pos + 1, self.block_size):
                fake_prev[p] = intermediate[p] ^ padding_value

            # 尝试当前字节的所有可能值（0x00 ~ 0xFF）
            found = False
            for guess in range(256):
                fake_prev[byte_pos] = guess
                test_ciphertext = bytes(fake_prev) + target_block                            #猜测的C_{i-1}和C_i拼接用于进行解密尝试

                if self.oracle.check(test_ciphertext):                                       #check没问题说明猜中了C_{i-1}的那一位
                    intermediate[byte_pos] = guess ^ padding_value                           #和目标填充值异或得到这一位的D_k(C_i)
                    plaintext[byte_pos] = intermediate[byte_pos] ^ prev_block[byte_pos]      #和真实C_{i-1}异或得到这一位的P(C_i)

                    if verbose:
                        char = chr(plaintext[byte_pos]) if 32 <= plaintext[byte_pos] < 127 else '?'
                        print(f"    ✓ guess=0x{guess:02x} → D(C)[{byte_pos}]=0x{intermediate[byte_pos]:02x} → P[{byte_pos}]=0x{plaintext[byte_pos]:02x} ('{char}')")
                    found = True
                    break

            if not found:
                raise RuntimeError(f"攻击第 {byte_pos} 个字节失败")

        if verbose:
            print(f"\n  恢复明文块: {bytes(plaintext).hex()}")
            try:
                unpadded = pkcs7_unpad(bytes(plaintext))
                print(f"  去填充后:   {unpadded.hex()} → '{unpadded.decode('utf-8', errors='replace')}'")
            except:
                print(f"  原始块:     '{bytes(plaintext).decode('utf-8', errors='replace')}'")

        return bytes(plaintext)

    def attack_full(self, iv, ciphertext, verbose=True):
        if len(ciphertext) % self.block_size != 0:
            raise ValueError("密文长度必须是16的整数倍")

        num_blocks = len(ciphertext) // self.block_size

        if verbose:
            print(f"\n{'#'*70}")
            print(f"# 开始完整攻击")
            print(f"# 密文块数: {num_blocks}")
            print(f"# IV: {iv.hex()}")
            print(f"# 密文: {ciphertext.hex()}")
            print(f"{'#'*70}")

        all_plaintext = bytearray()
        prev_block = iv  # 第一个块的前驱是 IV

        for i in range(num_blocks):
            target_block = ciphertext[i*self.block_size : (i+1)*self.block_size]
            block_plaintext = self.attack_block(prev_block, target_block, verbose)
            all_plaintext.extend(block_plaintext)
            prev_block = target_block

        # 去填充
        try:
            result = pkcs7_unpad(bytes(all_plaintext))
        except Exception as e:
            print(f"去填充失败: {e}")
            result = bytes(all_plaintext)

        if verbose:
            print(f"\n{'#'*70}")
            print(f"# 攻击完成！")
            print(f"# 恢复明文: {result.decode('utf-8', errors='replace')}")
            print(f"# Hex: {result.hex()}")
            print(f"{'#'*70}")

        return result


def demo_full_attack():
    """完整攻击流程——加密 → 攻击 → 恢复明文"""

    # 密钥和IV（16字节）
    KEY = bytes.fromhex("0123456789ABCDEFFEDCBA9876543210")
    IV  = bytes.fromhex("000102030405060708090A0B0C0D0E0F")

    # 原始明文
    original_plaintext = "Hello,SMB2026-CBC-Oracle-Padding-Attack"

    print("="*70)
    print("演示 1: 完整攻击流程")
    print("="*70)
    print(f"\n原始明文: '{original_plaintext}'")
    print(f"原始明文Hex: {original_plaintext.encode().hex()}")

    # Step 1: 加密
    padded_plaintext = pkcs7_pad(original_plaintext.encode())
    print(f"\n[Step 1] PKCS7填充后: {padded_plaintext.hex()}")

    ciphertext = sm4_cbc_encrypt(original_plaintext.encode(), KEY, IV)
    print(f"[Step 1] 加密后密文: {ciphertext.hex()}")
    print(f"         密文长度: {len(ciphertext)} 字节 ({len(ciphertext)//16} 个块)")

    # Step 2: 创建填充预言 Oracle
    print(f"\n[Step 2] 创建填充预言 Oracle（模拟服务端）")
    oracle = PaddingOracle(KEY, IV)

    # 验证 Oracle 工作正常
    print(f"         验证: 正确密文 -> {oracle.check(ciphertext)}")
    tampered = bytearray(ciphertext)
    tampered[-1] ^= 0xFF
    print(f"         验证: 篡改密文 -> {oracle.check(bytes(tampered))}")

    # Step 3: 执行攻击
    print(f"\n[Step 3] 开始填充预言攻击...")
    attacker = PaddingOracleAttacker(oracle)
    recovered = attacker.attack_full(IV, ciphertext, verbose=True)

    # Step 4: 验证结果
    print(f"\n[Step 4] 验证结果")
    print(f"原始明文: '{original_plaintext}'")
    print(f"恢复明文: '{recovered.decode('utf-8', errors='replace')}'")
    print(f"攻击成功: {original_plaintext.encode() == recovered}")

    return recovered



if __name__ == "__main__":
    demo_full_attack()