c - AES 混合列错误

Question

现在我正在学习密码学。所以，(为了练习和乐趣)，我决定实现 AES。我陷入了一点(这里是我的代码混合列):

typedef vector< vector<short> > vvector;

short mixBox[4][4] = 
{
    {0x02, 0x03, 0x01, 0x01},
    {0x01, 0x02, 0x03, 0x01},
    {0x01, 0x01, 0x02, 0x03},
    {0x03, 0x01, 0x01, 0x02}
};

short gfMultiply(short h1, short h2)
{
    //h1 can 0x01, 0x02 or 0x03
}

void mixColumns(vvector & v)
{
    vvector res(v.begin(), v.end());
    for(int i=0; i<4; i++)
        for(int j=0; j<4; j++)
            v[i][j] = 0x00;

    for(int i=0; i<4; i++)
        for(int j=0; j<4; j++)
            for(int k=0; k<4; k++)
                v[i][j] = v[i][j] ^ gfMultiply(mixBox[i][k], res[k][j]);
}

理论上，我理解乘法 gf(2^8)，但是对于实现算法，我有问题。我提到了这个site .但是要么我不明白某些点，要么我做错了什么。在维基百科中我读到了这个:

"The multiplication operation is defined as: multiplication by 1 means no change, multiplication by 2 means shifting to the left, and multiplication by 3 means shifting to the left and then performing xor with the initial unshifted value. After shifting, a conditional xor with 0x1B should be performed if the shifted value is larger than 0xFF."

假设上面我已经实现了这个:

short gfMultiply(short h1, short h2)
{
    //h1 can 0x01, 0x02 or 0x03
    short r;
    if(h1==0x01)
        return h2;
    if(h1==0x02)
        r = (h2<<1);
    else
        r = (h2<<1)^h2;
    if(r>0xFF)
        r = r^0x1b;
    return r;
}

但是当我测试时，结果不正确。我在这里做错了什么？

Answer 1

抱歉，有错误。我自己修好了，这是正确的:

short gfMultiply(short h1, short h2)
{
    //h1 can 0x01, 0x02 or 0x03
    short r;
    if(h1==0x01)
        return h2;
    if(h1==0x02)
    {
        r = (h2<<1);
        if(r>0xFF)
            r = r^0x11b;
    }
    else
    {
        r = (h2<<1);
        if(r>0xFF)
            r = r^0x11b;
        r = r^h2;
    }
    return r;
}