C# 與 Java Rsa加密與解密互通

Rsa 加密標准的制定已經過去了十多年了. 這兩天在看rsa 加密的文章,基本上都是在說 .net 與 java 之間的 rsa加密是不能互通的.因為項目有用到,所以花了點時間對rsa加密做了一點點了解,發現,不管是java 還是 C# 都對 rsa 的標准加密進行了實現, 是 對於標准是實現,不能互通就講不過去了. 今天特意寫了一段java 代碼試了一下,發現是完全可以的.

 

密鑰的描述: C#(.net) 中有三種方式導出密鑰,一種是blob,一種是 xml 另一種是參數,其中xml 的方式是把 參數進行了 xml序列化.blob 的形式我沒看懂是什么意思,只知道文檔上說是給微軟的什么api用的,下面給出其參數的形式.

RSAParameters   字段	Contains	對應的 PKCS #1 字段
D	d，私鑰指數	privateExponent
DP	d mod (p - 1)	exponent1
DQ	d mod (q - 1)	exponent2
Exponent	e，公鑰指數	publicExponent
InverseQ	(InverseQ)(q) = 1 mod p	coefficient
Modulus	n	modulus
P	p	prime1
Q	q	prime2

 

RSA 算法

若要生成密鑰對，可以從創建名為 p 和 q 的兩個大的質數開始; 這兩個數相乘，結果稱為 n; 因為 p 和 q 都是質數，所以 n 的全部因數為 1、p、q 和 n;

如果僅考慮小於 n 的數，則與 n 為互質數（即與 n 沒有公因數）的數的個數等於 (p - 1)(q - 1);

現在，選擇一個數 e，它與計算的值為互質數; 則公鑰表示為 {e, n};

若要創建私鑰，則必須計算 d，它是滿足 (d)(e) mod n = 1 的一個數; 根據 Euclidean 算法，私鑰為 {d, n};

純文本 m 到密碼文本 c 的加密定義為 c = (m ^ e) mod n; 解密則定義為 m = (c ^ d) mod n;

 

總之,我們可以從 .net 的rsa 中拿 到合適的密鑰描述就是了,至於 java的,我想也是可以做到的,而且通常密鑰是應該放在密鑰容器的.

 

接下來我們用.net 生成一個  rsa 的密鑰,以十六進制的方式輸出的(new 一個RSACryptoServiceProvider,把密鑰導出就可以了)

 

D: 2FE7479CF4CFEE63218C44D763C3E552DC5FBC94A31F944B88AE8E58F0ED16874B8BED35307B143F413761B2ECFFC95F48DF0D0A29FC155C0B968EFE9FFF36E7
DP: 6777B761BC29637622FC63682243BB2E05CCFC6FF710ADE1DCE6B0C843B17C4F
DQ: 68771CCDA40F0DA0B504C438BB03F7DF30F77364094D475E70270D148260D247
Exponent: 010001
InverseQ: 5665AB47697008CC2CECB544B582B9C50628281C400846C1E736629B03FE5C85
Modulus: B3F276C8EDF515FD3248CCF4163480B9F77443A666522D66B89411EC6DFE11DEA917A97C977750EE777DACBD4D2C11BC363FDC110E5CCA0A1361D51AFA4A7ADD
P: ECC60A01B1BDCBA1C5422D8A0A34FC0E46727DB4ED5089E54C356F052E0AB573
Q: C28F233948483D0CD0E3FA7B5D2955F2B15E831B38876FB0E7180D873EDF7A6F

 

 

為了方便寫.net 代碼,這里貼一下blob的密鑰

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

 

 

因為是私鑰,所以同時可以加密和解密.下面上  C# 代碼,比較簡單

 

using System;
    using System.Security.Cryptography;

    class Program
    {
        public static void Main(string[] args)
        {
            
            byte[] plainText = new byte[]{0,1,2,3,4,5};
            byte[] cipherText;
            byte[] key = String2Bytes("0702000000A40000525341328001000001000100D3D10816051881319774576B67B1D24F3AA303471A4402AB625208EC1CB04D508AF2098227C5EE185890ECB83E6971C12BDCF4F8AB0FD729167C815D3404C1AD1C0628E3544C89E9F9044A6869447310C72D7CDEB5E3582AB28BCC5069D60CD4AB5A1BB8C754AEB544FA65FB990ADB68F5AA37D7AC2CCDBF19058A2A4CF18FC29577D184E75EFD0ABEC1263893262F40C89AB2831B20B52B477770BA95FA02A71339911EBBF8630AD1AEECC205888440037D580B18AEF11FC5F35EB74E434E5A9302823BF795A34DADDFE0C6D55BC7997E667ADBE511BA06");
            using(var rsa = new RSACryptoServiceProvider())
            {
                rsa.ImportCspBlob(key);
                Console.WriteLine("OAEP:");
                Console.WriteLine(Bytes2String(rsa.Encrypt(plainText, true)));
                Console.WriteLine("PCSK1-v1_5:");
                Console.WriteLine(Bytes2String(rsa.Encrypt(plainText, false)));
            }
            Console.Write("Press any key to continue . . . ");
            Console.ReadKey(true);
        }
        
        
        const string pattern = @"[^0-9a-fA-F]+";
        static byte[] String2Bytes(string str)
        {
            str = System.Text.RegularExpressions.Regex.Replace(str, pattern, "");
            if (str == string.Empty)
                return null;
            byte[] data = new byte[str.Length / 2];
            
            for (int i = 0; i < data.Length; ++i)
                data[i] = byte.Parse(str.Substring(2 * i, 2), System.Globalization.NumberStyles.HexNumber);
            return data;
        }
        static string Bytes2String(byte[] data)
        {
            System.Text.StringBuilder builder = new System.Text.StringBuilder();
            foreach (var  element in data) {
                builder.AppendFormat("{0:X2}",element);
            }
            return builder.ToString();
        }
    }

運行后輸出(rsa加密,密文每次都不一定一樣):

OAEP:
87F04B0F28B81D23E63DA71C8278E0B7E357F40583BDDCAB493D44A58080EB178EC8E0DB0DCD4BE5427FDB8190229B8DF2511BDA1082607C92BD03B0615D5AD3
PCSK1-v1_5:
358AB4D336D0C35DAE3895E8A125F4F5AD0FB58117A4100FAF15DE95FF8615F01FFB1A59C9B579792B7C14E93E54A3E7E236D464DDB93D8DF9D96F63F46BACD7

現在我們有密文了,至於.net 的解密 .net 加密后的密文,我沒興趣去看,反正鐵定可以的就是了.

 

下面我們寫java 的解密部分

package rsatest;

import java.math.BigInteger;
import java.security.KeyFactory;
import java.security.PrivateKey;
import java.security.PublicKey;
import java.security.spec.RSAPrivateKeySpec;
import java.security.spec.RSAPublicKeySpec;
import javax.crypto.Cipher;

public class RsaTest {

    public static void main(String[] args) throws Exception {
        //前面補了個0,符號位為0,表示非負
        BigInteger n = new BigInteger("B3F276C8EDF515FD3248CCF4163480B9F77443A666522D66B89411EC6DFE11DEA917A97C977750EE777DACBD4D2C11BC363FDC110E5CCA0A1361D51AFA4A7ADD", 16);
        BigInteger e = new BigInteger("010001", 16);
        BigInteger d = new BigInteger("2FE7479CF4CFEE63218C44D763C3E552DC5FBC94A31F944B88AE8E58F0ED16874B8BED35307B143F413761B2ECFFC95F48DF0D0A29FC155C0B968EFE9FFF36E7", 16);
        RSAPublicKeySpec keySpec = new RSAPublicKeySpec(n, e);
        PublicKey publicKey = KeyFactory.getInstance("RSA").generatePublic(keySpec);
        RSAPrivateKeySpec prvKeySpec = new RSAPrivateKeySpec(n, d);
        PrivateKey privateKey = KeyFactory.getInstance("RSA").generatePrivate(prvKeySpec);

        //現在key 准備好了,把前面的密文放這里解密
        byte[] oaepCiphertext = Hex2Bytes("87F04B0F28B81D23E63DA71C8278E0B7E357F40583BDDCAB493D44A58080EB178EC8E0DB0DCD4BE5427FDB8190229B8DF2511BDA1082607C92BD03B0615D5AD3");

        //解密OAEP 加密的數據
        Cipher cipher = Cipher.getInstance("RSA/ECB/OAEPPADDING");
        cipher.init(Cipher.DECRYPT_MODE, privateKey);
        byte[] plaintext = cipher.doFinal(oaepCiphertext);
        System.out.println(Bytes2Hex(plaintext));

        //解密PKCS1-v1_5 加密的數據
        byte[] ciphertext = Hex2Bytes("358AB4D336D0C35DAE3895E8A125F4F5AD0FB58117A4100FAF15DE95FF8615F01FFB1A59C9B579792B7C14E93E54A3E7E236D464DDB93D8DF9D96F63F46BACD7");
        cipher = Cipher.getInstance("RSA");
        cipher.init(Cipher.DECRYPT_MODE, privateKey);
        plaintext = cipher.doFinal(ciphertext);
        System.out.println(Bytes2Hex(plaintext));

    }

    public static byte[] Hex2Bytes(String hexStr) {
        if (hexStr.length() % 2 != 0) {
            hexStr = "0" + hexStr;
        }
        byte[] bytes = new byte[hexStr.length() / 2];
        for (int i = 0; i < bytes.length; ++i) {
            bytes[i] = (byte) Integer.parseUnsignedInt(hexStr.substring(i * 2, i * 2 + 2), 16);
        }
        return bytes;
    }

    public static String Bytes2Hex(byte[] bytes) {
        StringBuilder builder = new StringBuilder();
        for (byte b : bytes) {
            builder.append(String.format("%02X", b));
        }
        return builder.toString();
    }
}

程序運行后輸出:

000102030405
000102030405

跟我們預期的是一樣的.

 

我們再用java 對明文加密,代碼片段

// plaintext {0,1,2,3,4,5}
        cipher = Cipher.getInstance("RSA/ECB/OAEPPADDING");
        cipher.init(Cipher.ENCRYPT_MODE, publicKey);
        oaepCiphertext = cipher.doFinal(plaintext);
        System.out.println("OAEP:");
        System.out.println(Bytes2Hex(oaepCiphertext));
        
        cipher = Cipher.getInstance("RSA");
        cipher.init(Cipher.ENCRYPT_MODE, publicKey);
        ciphertext = cipher.doFinal(plaintext);
        System.out.println("PCSK1-v1_5:");
        System.out.println(Bytes2Hex(ciphertext));

對應輸出(rsa加密,密文每次都不一定一樣):

OAEP:
3144C4CB06C7F49D31E65D09C840069F7CCF602487908CCEAB33D473B949199E1795530B69E1FA20EB59E392B2B934024D46E979DEA1682BDFA61D6FDD980F9C
PCSK1-v1_5:
699A694BEB75616879C6B8D311CC10D987EA109D494EE6C9380CD2C02A124613F130C440CB1CA6D3405E50B62CF96A79EB43C3370253E5D8C1A9132CFE01D686

 

接下來用C# 對數據解密,代碼片段

//還是用之前那個rsa對象
//OAEP
cipherText = String2Bytes("3144C4CB06C7F49D31E65D09C840069F7CCF602487908CCEAB33D473B949199E1795530B69E1FA20EB59E392B2B934024D46E979DEA1682BDFA61D6FDD980F9C");
Console.WriteLine(Bytes2String(rsa.Decrypt(cipherText, true)));
//PCSK1-v1_5
cipherText = String2Bytes("699A694BEB75616879C6B8D311CC10D987EA109D494EE6C9380CD2C02A124613F130C440CB1CA6D3405E50B62CF96A79EB43C3370253E5D8C1A9132CFE01D686");
Console.WriteLine(Bytes2String(rsa.Decrypt(cipherText, false)));

對應的輸出:

000102030405
000102030405

解密也成功了.

 

或許該頭疼的問題是,已知  {e,n} 或者  {d,n} 怎么生成.net 使用的密鑰.