SM4国密算法Java版

根据 国密SM4 文档 编写的一个Java 加密解密样例

package javasm4;
/**
 *
 * @author Jeen
 */
public class JavaSM4 {
          
    public static int[] key = new int[4];//密钥
    public static int[] temp = new int[4];//中间量 存储运算结果
    public static int[] rkey = new int[32];//轮密钥
    public static int[] fk = {0xa3b1bac6, 0x56AA3350, 0x677d9197, 0xb27022dc}; //系统参数
    public static int[] ck = {//ck[i][j] = (4i+j)×7（mod 256）   i=0->31   j=0->3 //固定参数
        0x00070e15, 0x1c232a31, 0x383f464d, 0x545b6269,
        0x70777e85, 0x8c939aa1, 0xa8afb6bd, 0xc4cbd2d9,
        0xe0e7eef5, 0xfc030a11, 0x181f262d, 0x343b4249,
        0x50575e65, 0x6c737a81, 0x888f969d, 0xa4abb2b9,
        0xc0c7ced5, 0xdce3eaf1, 0xf8ff060d, 0x141b2229,
        0x30373e45, 0x4c535a61, 0x686f767d, 0x848b9299,
        0xa0a7aeb5, 0xbcc3cad1, 0xd8dfe6ed, 0xf4fb0209,
        0x10171e25, 0x2c333a41, 0x484f565d, 0x646b7279};
    private static int[] sbi = { //sbi 用于8位置换的数组 sbox
        0xd6,0x90,0xe9,0xfe,0xcc,0xe1,0x3d,0xb7,0x16,0xb6,0x14,0xc2,0x28,0xfb,0x2c,0x05,
        0x2b,0x67,0x9a,0x76,0x2a,0xbe,0x04,0xc3,0xaa,0x44,0x13,0x26,0x49,0x86,0x06,0x99,
        0x9c,0x42,0x50,0xf4,0x91,0xef,0x98,0x7a,0x33,0x54,0x0b,0x43,0xed,0xcf,0xac,0x62,
        0xe4,0xb3,0x1c,0xa9,0xc9,0x08,0xe8,0x95,0x80,0xdf,0x94,0xfa,0x75,0x8f,0x3f,0xa6,
        0x47,0x07,0xa7,0xfc,0xf3,0x73,0x17,0xba,0x83,0x59,0x3c,0x19,0xe6,0x85,0x4f,0xa8,
        0x68,0x6b,0x81,0xb2,0x71,0x64,0xda,0x8b,0xf8,0xeb,0x0f,0x4b,0x70,0x56,0x9d,0x35,
        0x1e,0x24,0x0e,0x5e,0x63,0x58,0xd1,0xa2,0x25,0x22,0x7c,0x3b,0x01,0x21,0x78,0x87,
        0xd4,0x00,0x46,0x57,0x9f,0xd3,0x27,0x52,0x4c,0x36,0x02,0xe7,0xa0,0xc4,0xc8,0x9e,
        0xea,0xbf,0x8a,0xd2,0x40,0xc7,0x38,0xb5,0xa3,0xf7,0xf2,0xce,0xf9,0x61,0x15,0xa1,
        0xe0,0xae,0x5d,0xa4,0x9b,0x34,0x1a,0x55,0xad,0x93,0x32,0x30,0xf5,0x8c,0xb1,0xe3,
        0x1d,0xf6,0xe2,0x2e,0x82,0x66,0xca,0x60,0xc0,0x29,0x23,0xab,0x0d,0x53,0x4e,0x6f,
        0xd5,0xdb,0x37,0x45,0xde,0xfd,0x8e,0x2f,0x03,0xff,0x6a,0x72,0x6d,0x6c,0x5b,0x51,
        0x8d,0x1b,0xaf,0x92,0xbb,0xdd,0xbc,0x7f,0x11,0xd9,0x5c,0x41,0x1f,0x10,0x5a,0xd8,
        0x0a,0xc1,0x31,0x88,0xa5,0xcd,0x7b,0xbd,0x2d,0x74,0xd0,0x12,0xb8,0xe5,0xb4,0xb0,
        0x89,0x69,0x97,0x4a,0x0c,0x96,0x77,0x7e,0x65,0xb9,0xf1,0x09,0xc5,0x6e,0xc6,0x84,
        0x18,0xf0,0x7d,0xec,0x3a,0xdc,0x4d,0x20,0x79,0xee,0x5f,0x3e,0xd7,0xcb,0x39,0x48};
          
          
    public static void main(String[] args)
    {
        JavaSM4 sm = new JavaSM4();
        int[] msg = {0x01234567, 0x89abcdef, 0xfedcba98, 0x76543210};
        int[] smsg = {0x595298c7, 0xc6fd271f, 0x0402f804, 0xc33d3f66};
        key[0] = 0x01234567;
        key[1] = 0x89abcdef;
        key[2] = 0xfedcba98;
        key[3] = 0x76543210;
        int j=0,n=1000000;
        long startTime = System.currentTimeMillis();   //获取开始时间
        for(j=0; j<n; j++)
        {
            //msg = sm4(msg,1);//加密运算
            smsg = sm4(smsg,0);//解密运算
        }
        long endTime = System.currentTimeMillis(); //获取结束时间
        System.out.println("执行SM4加密运算"+n+"次时间： "+(endTime-startTime)+"ms");
    }
          
    private static int[] sm4(int[] t,int s) //s!=0 时为加密运算，s=0时为解密运算
    {
        rkey = initrk();
        if(s == 0)
        {
            rkey = r(rkey);
        }
        int[] x = new int[36];
        x[0] = t[0];
        x[1] = t[1];
        x[2] = t[2];
        x[3] = t[3];
        int i;
        for(i=0;i<32;i++)
        {
            x[i+4] = f(x[i],x[i+1],x[i+2],x[i+3],rkey[i]);
        }
        x = r(x);
        temp[0] = x[0];
        temp[1] = x[1];
        temp[2] = x[2];
        temp[3] = x[3];
        return temp;
    }
            
    private static int[] initrk()
    {
        int i;
        int[] k = new int[36];
        int[] rk = new int[32];
        k[0] = key[0] ^ fk[0];
        k[1] = key[1] ^ fk[1];
        k[2] = key[2] ^ fk[2];
        k[3] = key[3] ^ fk[3];
        for(i=0;i<32;i++)
        {
            rk[i] = k[i+4] = k[i] ^ tn(k[i+1]^k[i+2]^k[i+3]^ck[i]);
        }
        return rk;
    }
          
    private static int[] r(int[] x)
    {
        int[] t = new int[x.length];
        int i;
        for(i=0; i<x.length; i++)
        {
            t[i] = x[x.length - 1 -i];
        }
        return t;
    }
          
    private static int f(int x0,int x1,int x2,int x3,int k)
    {
        return (x0 ^ t(x1 ^ x2 ^ x3 ^ k));
    }
          
    private static int t(int ta)
    {
        return l(tj(ta));
    }
          
    private static int tn(int ta)
    {
        return ln(tj(ta));
    }
          
    private static int l(int temp)
    {
        return temp ^ Px(temp,2) ^ Px(temp,10) ^ Px(temp,18) ^ Px(temp,24);
    }
    private static int ln(int temp)
    {
        return temp ^ Px(temp,13) ^ Px(temp,23);
    }
          
    private static int tj(int a)
    {
        byte[] b = new byte[4];
        byte[] c = new byte[4];
        c = intToBytes(a);
        b[0] = sbox(c[0]);
        b[1] = sbox(c[1]);
        b[2] = sbox(c[2]);
        b[3] = sbox(c[3]);
        a = bytesToInt(b[0],b[1],b[2],b[3]);
        return a;
    }
          
    private static byte sbox(byte a) //S盒 8 bit 置换
    { 
        int t = (a << 24) >>> 24;
        return (byte)sbi[t];
    }
          
    private static int Px(int x,int n) //整型循环左移运算 n <=32 //java中没有无符号整型,需要注意移位后的填充符
    {
        return ((x<<n)|(x>>>(32-n)));
    }
    private static int bytesToInt(byte b0,byte b1,byte b2,byte b3) // int = 4 * byte = 32 bit unsigned
    {
        int tint = 0;
        int temp = b0 << 24;
        tint = temp;
        temp = (b1 << 24) >>> 8;
        tint |= temp;
        temp = (b2 << 24) >>> 16;
        tint |= temp;
        temp = (b3 << 24) >>> 24;
        tint |= temp;
        return tint;
    }
    private static byte[] intToBytes(int i)
    {
        byte[] tbyte = new byte[4];
        tbyte[0] = (byte)(i >>> 24);
        tbyte[1] = (byte)((i<<8)>>>24);
        tbyte[2] = (byte)((i<<16)>>>24);
        tbyte[3] = (byte)((i<<24)>>>24);
        return tbyte;
    }
}

TODO： 1文件信息流处理   2长度不足的填充算法