我想将RSA私钥分为两半,并将它们存储在两个不同的地方,我该怎么办?
public GenerateKeys(int keylength) throws NoSuchAlgorithmException, NoSuchProviderException {
keylength=512;
this.keyGen = KeyPairGenerator.getInstance("RSA");
SecureRandom random = SecureRandom.getInstance("SHA1PRNG");
this.keyGen.initialize(keylength, random);
}
最佳答案
这是一个示例,它将您的私钥分为两个部分,D1和D2。类似于here的讨论
import java.security.KeyPair;
import java.security.KeyFactory;
import java.security.KeyPairGenerator;
import java.security.PrivateKey;
import java.security.PublicKey;
import java.security.NoSuchAlgorithmException;
import java.security.SecureRandom;
import java.security.spec.EncodedKeySpec;
import java.security.spec.PKCS8EncodedKeySpec;
import java.security.spec.X509EncodedKeySpec;
public class OnetimePad{
public static byte[] xor(byte[] key, byte[] rand){
if(key.length != rand.length){
return null;
}
byte[] ret = new byte[key.length];
for(int i =0; i < key.length; i++){
ret[i] = (byte)((key[i] ^ rand[i]) );
}
return ret;
}
public static void main(String []args) throws Exception{
SecureRandom random = new SecureRandom();
KeyPairGenerator keyGen = KeyPairGenerator.getInstance("RSA");
keyGen.initialize(1024);
KeyPair keypair = keyGen.genKeyPair();
PrivateKey privateKey = keypair.getPrivate();
byte[] privateKeyBytes = privateKey.getEncoded();
//Private Key Part 1
byte[] D1 = new byte[privateKeyBytes.length];
random.nextBytes(D1);
//Private Key Part 2
byte[] D2 = xor(privateKeyBytes, D1);
//now D1 and D2 are split parts of private keys..
//Let's verify if we could reproduce them back
byte[] privateKeyByesTmp = xor(D2, D1);
KeyFactory keyFactory = KeyFactory.getInstance("RSA");
EncodedKeySpec privateKeySpec = new PKCS8EncodedKeySpec(privateKeyByesTmp);
PrivateKey privateKey2 = keyFactory.generatePrivate(privateKeySpec);
boolean same = privateKey.equals(privateKey2);
if(same){
System.out.println("Key loaded successfully");
}else{
System.out.println("Ooops");
}
}
}
注意:
请检查random seed上SecureRandom的以下文档。特别强调的部分
许多SecureRandom实现都采用伪随机数生成器(PRNG)的形式,这意味着它们使用确定性算法从真实的随机种子生成伪随机序列。其他实现可能会产生真正的随机数,而其他实现可能会同时使用这两种技术。