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我正在尝试自己编码 SHA3-512。我坚持 theta 并且我无法根据这个测试向量从中获得正确的输出。消息只有 200 个字节的空字节，我填充了消息。

输入数组是：

[0, 0] = 0000000000000006 
[1, 0] = 0000000000000000 
[2, 0] = 0000000000000000 
[3, 0] = 0000000000000000 
[4, 0] = 0000000000000000 
[0, 1] = 0000000000000000 
[1, 1] = 0000000000000000 
[2, 1] = 0000000000000000 
[3, 1] = 8000000000000000 
[4, 1] = 0000000000000000 
[0, 2] = 0000000000000000 
[1, 2] = 0000000000000000 
[2, 2] = 0000000000000000 
[3, 2] = 0000000000000000 
[4, 2] = 0000000000000000 
[0, 3] = 0000000000000000 
[1, 3] = 0000000000000000 
[2, 3] = 0000000000000000 
[3, 3] = 0000000000000000 
[4, 3] = 0000000000000000 
[0, 4] = 0000000000000000 
[1, 4] = 0000000000000000 
[2, 4] = 0000000000000000 
[3, 4] = 0000000000000000 
[4, 4] = 0000000000000000

我在 C 中的 theta 代码：

    /* θ */

    uint64_t A[5] = {0x00};
    uint64_t B[5] = {0x00};

    for (i = 0; i < 5; i++){
        A[i] = square[i][0] ^ square[i][1] ^ square[i][2] ^ square[i][3] ^ square[i][4];
    }

    for (i = 0; i < 5; i++){
        B[i] = A[(i + 4) % 5] ^ rotate_left_64(A[(i + 1) % 5], 1);
        for (j = 0; j < 5; j++){
            square[i][j] = square[i][j] ^ B[i];
        }
    }

函数 rotate_left_64 ：

uint64_t rotate_left_64(uint64_t i, int c){
    uint64_t j = 0;
    j = i << c;
    j = j ^ (i >> (64 - c));
    return j;
}

现在工作了几个小时......我的错误是什么？

提前致谢

4

1 回答 1

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好的。我想到了。代码应该是这样的：

        i = 0;
        for (j = 0; j < 5; j++){
            A[i] = square[0][j] ^ square[1][j] ^ square[2][j] ^ square[3][j] ^ square[4][j];
            i++;
        }

        for (i = 0; i < 5; i++){
            B[i] = A[(i + 4) % 5] ^ rotate_left_64(A[(i + 1) % 5], 1);
        }

        for (j = 0; j < 5; j++){
            for (i = 0; i < 5; i++){
                square[i][j] = square[i][j] ^ B[j];
            }
        }
于 2022-02-10T11:05:29.420 回答