java - 为什么用AES加密16个字节的密文是32个字节？

Question

我使用加密AES算法，当我加密16字节(一个 block )时，结果是32字节。这样可以吗？

我使用的源代码是:

package net.sf.andhsli.hotspotlogin;

import java.security.SecureRandom;

import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.SecretKey;
import javax.crypto.spec.SecretKeySpec;

/**
 * Usage:
 * <pre>
 * String crypto = SimpleCrypto.encrypt(masterpassword, cleartext)
 * ...
 * String cleartext = SimpleCrypto.decrypt(masterpassword, crypto)
 * </pre>
 * @author ferenc.hechler
 */
public class SimpleCrypto {

    public static String encrypt(String seed, String cleartext) throws Exception {
        byte[] rawKey = getRawKey(seed.getBytes());
        byte[] result = encrypt(rawKey, cleartext.getBytes());
        return toHex(result);
    }

    public static String decrypt(String seed, String encrypted) throws Exception {
        byte[] rawKey = getRawKey(seed.getBytes());
        byte[] enc = toByte(encrypted);
        byte[] result = decrypt(rawKey, enc);
        return new String(result);
    }

    private static byte[] getRawKey(byte[] seed) throws Exception {
        KeyGenerator kgen = KeyGenerator.getInstance("AES");
        SecureRandom sr = SecureRandom.getInstance("SHA1PRNG");
        sr.setSeed(seed);
        kgen.init(128, sr); // 192 and 256 bits may not be available
        SecretKey skey = kgen.generateKey();
        byte[] raw = skey.getEncoded();
        return raw;
    }


    private static byte[] encrypt(byte[] raw, byte[] clear) throws Exception {
        SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
        Cipher cipher = Cipher.getInstance("AES");
        cipher.init(Cipher.ENCRYPT_MODE, skeySpec);
        byte[] encrypted = cipher.doFinal(clear);
        return encrypted;
    }

    private static byte[] decrypt(byte[] raw, byte[] encrypted) throws Exception {
        SecretKeySpec skeySpec = new SecretKeySpec(raw, "AES");
        Cipher cipher = Cipher.getInstance("AES");
        cipher.init(Cipher.DECRYPT_MODE, skeySpec);
        byte[] decrypted = cipher.doFinal(encrypted);
        return decrypted;
    }

    public static String toHex(String txt) {
        return toHex(txt.getBytes());
    }
    public static String fromHex(String hex) {
        return new String(toByte(hex));
    }

    public static byte[] toByte(String hexString) {
        int len = hexString.length()/2;
        byte[] result = new byte[len];
        for (int i = 0; i < len; i++)
            result[i] = Integer.valueOf(hexString.substring(2*i, 2*i+2), 16).byteValue();
        return result;
    }

    public static String toHex(byte[] buf) {
        if (buf == null)
            return "";
        StringBuffer result = new StringBuffer(2*buf.length);
        for (int i = 0; i < buf.length; i++) {
            appendHex(result, buf[i]);
        }
        return result.toString();
    }
    private final static String HEX = "0123456789ABCDEF";
    private static void appendHex(StringBuffer sb, byte b) {
        sb.append(HEX.charAt((b>>4)&0x0f)).append(HEX.charAt(b&0x0f));
    }

}

Answer 1

如果您查看 specification section 5然后你可以看到输入、输出和状态都是128位的。唯一不同的是 key 的大小:128、196 或 256 位。因此，加密 16 字节的输入状态将产生 16 字节的输出状态。

您确定没有将它与十六进制表示法或类似的长度混淆吗？如果它是十六进制表示法，那么它是正确的，因为每个字节都需要两个字符来表示它:00-FF(范围为 0-255)。因此，例如，16 个字节将被编码为 32 个十六进制字符。

另一种测试加密是否正确的方法是进行等效的解密，看看它是否与明文输入字符串匹配。

无论如何，它做了正确的事情。这是一个测试:

public static void main(String[] args) {
  try {
    String plaintext = "Hello world", key = "test";
    String ciphertext = encrypt(key, plaintext);
    String plaintext2 = decrypt(key, ciphertext);
    System.out.println("Encrypting '" + plaintext +
                       "' yields: (" + ciphertext.length() + ") " + ciphertext);
    System.out.println("Decrypting it yields: " + plaintext2);
  }
  catch (Exception ex) {
      ex.printStackTrace();
  }
}

产生:

Encrypting 'Hello world' yields: (32)5B68978D821FCA6022D4B90081F76B4F

Decrypting it yields: Hello world