PBEKeySpec与字节数组参数，而不是ASCII(PBEKeySpec with byte ar

我想知道是否有使用方式PBEKeySpec字节数组参数。

请找一个链接到以下文件：

http://docs.oracle.com/javase/1.7/docs/api/javax/crypto/spec/PBEKeySpec.html ）

Answer 1:

这下面是我的解决方案：我得到了它周围的Googling。请考虑我必须在内部复制密码和盐，因为它们具有另一种格式时，他们来自外部，但结果是一样的。看来它的工作原理，解决了有密码的byte []的问题，而不是为char []（这是我都快疯了），我希望它能帮助！干杯，Soosta

public class Pbkdf2 {

    public Pbkdf2() {
    }

    public void GenerateKey(final byte[] masterPassword, int masterPasswordLen,
                            final byte[] salt, int saltLen,
                            int iterationCount, int requestedKeyLen,
                            byte[] generatedKey) {

        byte[] masterPasswordInternal = new byte[masterPasswordLen];
        System.arraycopy(masterPassword, 0, masterPasswordInternal, 0, masterPasswordLen);
        byte[] saltInternal = new byte[saltLen];
        System.arraycopy(salt, 0, saltInternal, 0, saltLen);


        SecretKeySpec keyspec = new SecretKeySpec(masterPasswordInternal, "HmacSHA1");
        Mac prf = null;
        try {
            prf = Mac.getInstance("HmacSHA1");
            prf.init(keyspec);
        } catch (NoSuchAlgorithmException e) {
            e.printStackTrace();
        } catch (InvalidKeyException e) {
            e.printStackTrace();
        }

        int hLen = prf.getMacLength();   // 20 for SHA1
        int l = Math.max(requestedKeyLen, hLen); //  1 for 128bit (16-byte) keys
        int r = requestedKeyLen - (l - 1) * hLen;      // 16 for 128bit (16-byte) keys
        byte T[] = new byte[l * hLen];
        int ti_offset = 0;
        for (int i = 1; i <= l; i++) {
            F(T, ti_offset, prf, saltInternal, iterationCount, i);
            ti_offset += hLen;
        }

        System.arraycopy(T, 0, generatedKey, 0, requestedKeyLen);
    }

    private static void F(byte[] dest, int offset, Mac prf, byte[] S, int c, int blockIndex) {
        final int hLen = prf.getMacLength();
        byte U_r[] = new byte[hLen];
        // U0 = S || INT (i);
        byte U_i[] = new byte[S.length + 4];
        System.arraycopy(S, 0, U_i, 0, S.length);
        INT(U_i, S.length, blockIndex);
        for (int i = 0; i < c; i++) {
            U_i = prf.doFinal(U_i);
            xor(U_r, U_i);
        }

        System.arraycopy(U_r, 0, dest, offset, hLen);
    }

    private static void xor(byte[] dest, byte[] src) {
        for (int i = 0; i < dest.length; i++) {
            dest[i] ^= src[i];
        }
    }

    private static void INT(byte[] dest, int offset, int i) {
        dest[offset + 0] = (byte) (i / (256 * 256 * 256));
        dest[offset + 1] = (byte) (i / (256 * 256));
        dest[offset + 2] = (byte) (i / (256));
        dest[offset + 3] = (byte) (i);
    }
}

Answer 2:

我不得不实施一两相PBKDF2推导（因此第二PBKDF2有从所述第一输入字节）。我结束了使用BouncyCastle的，因为我只是无法得到的字节数组字符数组体操工作。感谢帕西从这个其他问题：可靠的实现PBKDF2-HMAC-SHA256为JAVA

import org.bouncycastle.crypto.generators.PKCS5S2ParametersGenerator;
import org.bouncycastle.crypto.digests.SHA256Digest;
import org.bouncycastle.crypto.digests.GeneralDigest;
import org.bouncycastle.crypto.params.KeyParameter;

GeneraDigest algorithm = new SHA256Digest();
PKCS5S2ParametersGenerator gen = new PKCS5S2ParametersGenerator(algorithm);
gen.init(passwordBytes, salt, iterations);
byte[] dk = ((KeyParameter) gen.generateDerivedParameters(256)).getKey();