想要使用 32 字节的 AES 256 CBC，但它显示 java.security.InvalidAlgorithmParameterException
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Want to use AES 256 CBC with 32 bytes but it shows java.security.InvalidAlgorithmParameterException
提问by Vaibs
I am using AES 256 CBC. I have 32 bytes of IV. But when i run this it shows an exception as:
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：
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
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:
我不知道如何解决这个问题。我搜索过，但我不知道如何解决这个问题。我第一次尝试安全概念。我的 AES 256 CBC 代码是：
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;
}
Can you help me that by changing what in this code i will be able to use 32 bytes of IV. Thanks in advance
你能帮我改变这段代码中的内容，我将能够使用 32 字节的 IV。提前致谢
Edit: My main function to which calls this functions:
编辑：我调用这个函数的主函数：
AESUtil.setENCRYPTION_KEY("96161d7958c29a943a6537901ff0e913efaad15bd5e7c566f047412179504ffb");
AESUtil.setENCRYPTION_IV("d41361ed2399251f535e65f84a8f1c57");
String decrypted = AESUtil.decrypt(new String(sw0SrUIKe0DmS7sRd9+XMgtYg+BUiAfiOsdMw/Lo2RA=)); // AES Decrypt
回答by Duncan Jones
The AES algorithm has a 128-bit block size, regardless of whether you key length is 256, 192 or 128 bits.
AES 算法的块大小为 128 位，无论您的密钥长度是 256、192 还是 128 位。
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.
当对称密码模式需要 IV 时，IV 的长度必须等于密码的块大小。因此，您必须始终对 AES 使用 128 位（16 字节）的 IV。
There is no way to use a 32 byte IV with AES.
没有办法在 AES 中使用 32 字节的 IV。
