RSA私钥解密使用充气城堡给出“InvalidCipherTextException”

Question

我正在为 j2me 创建一个使用充气城堡的加密/解密方法。我已经使用 openssl 和硬编码在代码中创建了私钥/公钥。加密工作正常（我得到加密字符串），但是当我尝试解密该密文时，我得到以下异常

org.bouncycastle.crypto.InvalidCipherTextException: unknown block type
 - org.bouncycastle.crypto.encodings.PKCS1Encoding.decodeBlock(PKCS1Encoding.java:362)
 - org.bouncycastle.crypto.encodings.PKCS1Encoding.processBlock(PKCS1Encoding.java:166)
 - com.ust.CryptoJ2me.RSADecrypt(CryptoJ2me.java:564)

PFB代码sn -p

public byte[] RSAEncrypt(byte[] toEncrypt) throws Exception {
    String cipherStr;
    String modStr ="AA878F0D9........";
    String expStr = "10001";
    BigInteger modulus = new BigInteger(modStr,16);
    BigInteger exponent = new BigInteger(expStr,16);
    System.out.println("modulus = "+modulus+"// exponent = "+exponent);
    RSApubKey = new RSAKeyParameters(false,modulus,exponent);

    if (RSApubKey == null)
        throw new Exception("Generate RSA keys first!");

    AsymmetricBlockCipher eng = new RSAEngine();
    eng = new PKCS1Encoding(eng);
    eng.init(true, RSApubKey);
    byte[] cipherByte = eng.processBlock(toEncrypt, 0, toEncrypt.length);
    return cipherByte;//cipherStr;
}

public String RSADecrypt (byte[] toDecrypt) throws Exception {
    System.out.println("toDecrypt = "+toDecrypt);
    //byte[] toDecByte = toDecrypt.getBytes("UTF-8");
    //System.out.println("toDecByte ="+toDecByte);
    String plainText;
    BigInteger RSAmod = new BigInteger("00d1aec38b8d189a0a1..",16);
    BigInteger RSAprivExp  = new BigInteger("2d7af1b1283688dadc16..",16);
    BigInteger RSApubExp = new BigInteger("10001",16);
    BigInteger RSAdp = new BigInteger("00f5847cc67ea018f10f16..",16);
    BigInteger RSAdq = new BigInteger("00daa299bf356c6c6db6a21..",16);
    BigInteger RSAp = new BigInteger("00e28dd601e878dd6b1c0c..",16);
    BigInteger RSAq  = new BigInteger("400ff2e2df018507e4c2be6..",16);
    BigInteger RSAqInv   = new BigInteger("00cf4b2ba101efb2378aee..",16);
    RSAPrivateCrtKeyParameters RSAprivKey = new RSAPrivateCrtKeyParameters(RSAmod, RSApubExp,
            RSAprivExp, RSAp, RSAq, RSAdp, RSAdq, RSAqInv);

  AsymmetricBlockCipher eng = new RSAEngine();

  eng = new PKCS1Encoding(eng);

  eng.init(false, RSAprivKey);
  byte[] plainByte = eng.processBlock(toDecrypt, 0, toDecrypt.length);
  plainText = new String(plainByte);

  return plainText;
}

Answer 1

公钥和私钥的模数不匹配。它应该匹配。私钥模数可能更小（假设大端符号和相同长度的十六进制字符串），这就是您收到此错误消息的原因。由于私钥包含模数和公共指数分量，您可以在加密代码中替换它们，看看解密是否适用于新的密文。

RSA 的工作原理是将某些消息提高到 e（公共指数）模m 的幂。这导致密文小于m，但接近它。解密通过将密文提高到 d（私有指数）模 m（相同的模数）的幂。

如果私钥模数远小于公钥模数，则密文会更大，无法解密。从数学上讲，可以“解密”密文，但您不会取回原始明文。图书馆可能会就此发出警告。