想要使用带有32字节的AES 256 CBC，但它显示了java.security.InvalidAlgorithmParameterException [英] Want to use AES 256 CBC with 32 bytes but it shows java.security.InvalidAlgorithmParameterException

问题描述

我正在使用AES 256 CBC。我有32个字节的IV。但是当我运行它时它显示一个异常：

I am using AES 256 CBC. I have 32 bytes of IV. But when i run this it shows an exception as:

Exception in thread "main" java.lang.RuntimeException: java.security.InvalidAlgorithmParameterException: Wrong IV length: must be 16 bytes long
    at com.abc.aes265cbc.AESUtil.decrypt(AESUtil.java:50)
    at com.abc.aes265cbc.Security.main(Security.java:48)
Caused by: java.security.InvalidAlgorithmParameterException: Wrong IV length: must be 16 bytes long
    at com.sun.crypto.provider.CipherCore.init(CipherCore.java:430)
    at com.sun.crypto.provider.AESCipher.engineInit(AESCipher.java:217)
    at javax.crypto.Cipher.implInit(Cipher.java:790)
    at javax.crypto.Cipher.chooseProvider(Cipher.java:848)
    at javax.crypto.Cipher.init(Cipher.java:1347)
    at javax.crypto.Cipher.init(Cipher.java:1281)
    at com.abc.aes265cbc.AESUtil.decrypt(AESUtil.java:47)
    ... 1 more

我不知道如何解决这个问题。我搜索了但是我没有得到如何解决这个问题。我是第一次尝试安全概念。
AES 256 CBC的代码是：

I don't know how to solve this. I searched but I am not getting how to solve this. I am trying security concepts for the first time. My code for the AES 256 CBC is:

 public static void setENCRYPTION_IV(String ENCRYPTION_IV) {
        AESUtil.ENCRYPTION_IV  =   ENCRYPTION_IV;
    }

    public static void setENCRYPTION_KEY(String ENCRYPTION_KEY) {
        AESUtil.ENCRYPTION_KEY  =   ENCRYPTION_KEY;
    }



    public static String encrypt(String src) {
        try {
            Cipher cipher = Cipher.getInstance("AES/CBC/PKCS5Padding");
            cipher.init(Cipher.ENCRYPT_MODE, makeKey(), makeIv());
            return Base64.encodeBytes(cipher.doFinal(src.getBytes()));
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
    }

    public static String decrypt(String src) {
        String decrypted = "";
        try {
            Cipher cipher = Cipher.getInstance("AES/CBC/PKCS5Padding");
            cipher.init(Cipher.DECRYPT_MODE, makeKey(), makeIv());
            decrypted = new String(cipher.doFinal(Base64.decode(src)));
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
        return decrypted;
    }

    static AlgorithmParameterSpec makeIv() {
        try {
            return new IvParameterSpec(ENCRYPTION_IV.getBytes("UTF-8"));
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }
        return null;
    }

    static Key makeKey() {
        try {
            MessageDigest md = MessageDigest.getInstance("SHA-256");
            byte[] key = md.digest(ENCRYPTION_KEY.getBytes("UTF-8"));
            return new SecretKeySpec(key, "AES");
        } catch (NoSuchAlgorithmException e) {
            e.printStackTrace();
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }

        return null;
    }

你可以通过改变这段代码中的内容来帮助我吗？使用32个字节的IV。
提前致谢

Can you help me that by changing what in this code i will be able to use 32 bytes of IV. Thanks in advance

编辑：
我调用此功能的主函数：

My main function to which calls this functions:

 AESUtil.setENCRYPTION_KEY("96161d7958c29a943a6537901ff0e913efaad15bd5e7c566f047412179504ffb");

    AESUtil.setENCRYPTION_IV("d41361ed2399251f535e65f84a8f1c57");
    String decrypted = AESUtil.decrypt(new String(sw0SrUIKe0DmS7sRd9+XMgtYg+BUiAfiOsdMw/Lo2RA=));   // AES Decrypt

推荐答案

AES算法有128-位块大小，无论密钥长度是256,192还是128位。

The AES algorithm has a 128-bit block size, regardless of whether you key length is 256, 192 or 128 bits.

当对称密码模式需要IV时，IV的长度必须相等到密码的块大小。因此，您必须始终使用带有AES的128位（16字节）IV。

When a symmetric cipher mode requires an IV, the length of the IV must be equal to the block size of the cipher. Hence, you must always use an IV of 128 bits (16 bytes) with AES.

无法使用带有AES的32字节IV。

There is no way to use a 32 byte IV with AES.

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