GCC-6.3.0中的奇怪行为 _大数据知识库

我目前正在用C实现数据加密标准（DES）。
DES使用多个置换函数，所有这些都是通过对条目值、置换表和置换表的大小调用permute来完成的。它根据置换表的条目返回条目的置换位（我理解C数组的方式，无论是静态的还是动态的，都是它们对应于内存中的地址）。
加密/解密过程是直接的，因为两者都是通过调用相同的函数来完成的（这里：des_crypt），并反转DES的密钥调度表给出的子密钥（这里：des_ksa）。在des_crypt中，有一个对f(r, k)的调用，它对应于上面定义的密码函数。
在调试这段代码时（主要是让它运行），我观察到它会“崩溃”。所谓“crash”，我的意思是在打印调试信息之前代码流会停止：我猜是段故障造成的
奇怪的部分来了（更像是废话）。
代码使用gcc (MinGW.org GCC-6.3.0-1) 6.3.0编译。执行后，似乎在f中执行k ^= permute(r, E, 48)时，permute表的条目无法访问（执行printf("%lu", t[i]);会导致程序突然结束，而不会打印数组的第i个元素），即假设的segfault。
但似乎gcc成功地编译了许多（@JohnBollinger，@adabsurdum，...）。因此，我找到了一个在线编译器：程序编译并成功执行。
在以前的版本中，我也犯了多个错误：宏缺少括号，我在swap函数中有一些未定义的行为，由于我对表缺乏理解，permute函数没有很好地执行。
通过重复编译和执行具有这些更改的程序来纠正这一切，产生了正确的结果。
然而，即使我现在能够在前一台机器上编译和执行程序，它也会产生错误的结果：可能是因为转换问题我确实认为这件事不重要，并认为它已经解决了。
我也会相应地更新标题。
下面是完整的代码：

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
#include <math.h>
#include <string.h>

#define ROTL(x, l, w) ((((x) << (l)) | ((x) >> ((w) - (l)))) & (1 << (w)) - (l))
#define ROTR(x, l, w) ((((x) >> (l)) | ((x) << ((w) - (l)))) & (1 << (w)) - (l))

void swap(void* a, void* b, size_t size)
{
    char temp;
    
    if(!a || !b)
        return;
    while(size)
    {
        temp = *(char*)a;
        *(char*)a = *(char*)b;
        *(char*)b = temp;
        a = (void*)((char*)a + 1); // As void pointer increment is a non portable extension of GCC.
        b = (void*)((char*)b + 1);
        size--;
    }
    
    return;
}

//----- DES -----

#define DES_ROUNDS 16

const uint8_t S1[64] = {
    14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
    0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
    4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
    15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13
};

const uint8_t S2[64] = {
    15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
    3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
    0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
    13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9
};

const uint8_t S3[64] = {
    10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
    13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
    13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
    1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12
};

const uint8_t S4[64] = {
    7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
    13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
    10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
    3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14
};

const uint8_t S5[64] = {
    2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
    14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
    4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
    11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3
};

const uint8_t S6[64] = {
    12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
    10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
    9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
    4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13
};

const uint8_t S7[64] = {
    4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
    13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
    1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
    6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12
};

const uint8_t S8[64] = {
    13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,
    1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
    7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
    2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11
};

uint8_t s(box, block) // 6-bit block, box is S[1-8]
uint8_t *box, block;
{
    
    block &= 0x3F;
    
    return box[((block & 0b11) << 4) + ((block & 0b111100) >> 2)];
}

// All substitution boxes are false.
const uint8_t IP[64] = {
    57, 49, 41, 33, 25, 17, 9, 1,
    59, 51, 43, 35, 27, 19, 11, 3,
    61, 53, 45, 37, 29, 21, 13, 5,
    63, 55, 47, 39, 31, 23, 15, 7,
    56, 48, 40, 32, 24, 16, 8, 0,
    58, 50, 42, 34, 26, 18, 10, 2,
    60, 52, 44, 36, 28, 20, 12, 4,
    62, 54, 46, 38, 30, 22, 14, 6
};

const uint8_t FP[64] = {
    39, 7, 47, 15, 55, 23, 63, 31,
    38, 6, 46, 14, 54, 22, 62, 30,
    37, 5, 45, 13, 53, 21, 61, 29,
    36, 4, 44, 12, 52, 20, 60, 28,
    35, 3, 43, 11, 51, 19, 59, 27,
    34, 2, 42, 10, 50, 18, 58, 26,
    33, 1, 41, 9, 49, 17, 57, 25,
    32, 0, 40, 8, 48, 16, 56, 24
};

const uint8_t E[48] = {
    31, 0, 1, 2, 3, 4,
    3, 4, 5, 6, 7, 8,
    7, 8, 9, 10, 11, 12,
    11, 12, 13, 14, 15, 16,
    15, 16, 17, 18, 19, 20,
    19, 20, 21, 22, 23, 24,
    23, 24, 25, 26, 27, 28,
    27, 28, 29, 30, 31, 0
};

const uint8_t P[32] = {
    15, 6, 19, 20,
    28, 11, 27, 16,
    0, 14, 22, 25,
    4, 17, 30, 9,
    1, 7, 23, 13,
    31, 26, 2, 8,
    18, 12, 29, 5,
    21, 10, 3, 24
};

const uint8_t PC1C[28] = {
    56, 48, 40, 32, 24, 16, 8,
    0, 57, 49, 41, 33, 25, 17,
    9, 1, 58, 50, 42, 34, 26,
    18, 10, 2, 59, 51, 43, 35
};

const uint8_t PC1D[28] = {
    62, 54, 46, 38, 30, 22, 14,
    6, 61, 53, 45, 37, 29, 21,
    13, 5, 60, 52, 44, 36, 28,
    20, 12, 4, 27, 19, 11, 3
};

const uint8_t PC2[48] = {
    13, 16, 10, 23, 0, 4,
    2, 27, 14, 5, 20, 9,
    22, 18, 11, 3, 25, 7,
    15, 6, 26, 19, 12, 1,
    40, 51, 30, 36, 46, 54,
    29, 39, 50, 44, 32, 47,
    43, 48, 38, 55, 33, 52,
    45, 41, 49, 35, 28, 31
};

uint8_t getls(i) // gives the number of left shifts needed based on the iterno, quickly done, should find faster alternatives
uint8_t i;
{
    uint8_t p;
    
    if(i == 1)
        return 1;
    p = (i - 2) / 7;
    if((int)(7 * p + 2) < (int)i)
        return 2;
    
    return 1;
}

uint64_t permute(x, t, s)
uint64_t x;
uint8_t *t, s;
{
    uint8_t i;
    uint64_t r;
    
    // i-th bit of input is the t[i]-th bit of output
    r = 0;
    for(i = 0; i < s; i++)
        r |= ((x >> i) & 0b1) << t[i]; // 0b1 Kenobi!
    
    return r;
}

uint32_t f(r, k)
uint32_t r;
uint64_t k;
{
    uint32_t temp;
    uint8_t i;
    const uint8_t *S[8] = {
        S1, S2, S3, S4,
        S5, S6, S7, S8
    };
    
    k ^= permute(r, E, 48); // Doesn't get past this point.
    temp = 0;
    for(i = 0; i < 8; i++)
    {
        temp |= s(S[i], (k >> ((7 - k) * 6)) & 0b111111) & 0b1111;
        temp <<= 4;
    }
    
    return (uint32_t)permute(temp, P, 32);
}

uint64_t* des_ksa(key, rounds)
uint64_t key;
uint8_t rounds;
{
    uint64_t *subkeys;
    uint32_t c, d;
    uint8_t i, l;
    
    rounds &= 0x7F; // Make sure there are no overflows.
    subkeys = (uint64_t*)malloc(rounds * sizeof(uint64_t));
    if(!subkeys)
        return NULL;
    c = permute(key, PC1C, 28);
    d = permute(key, PC1D, 28);
    for(i = 0; i < rounds; i++)
    {
        l = getls(i);
        c = ROTL(c, l, 28);
        d = ROTL(d, l, 28);
        subkeys[i] = permute(((uint64_t)c << 28) | d, PC2, 48);
    }
    
    return subkeys;
}

/*
    :en: keep the subkeys' order as given by the des_ksa();
    :de: reverse the subkeys' order (relative to the one given by the des_ksa());
*/
uint64_t des_crypt(subkeys, block, rounds)
uint64_t *subkeys, block;
uint8_t rounds;
{
    uint32_t l, r, copy;
    uint8_t i;
    
    rounds &= 0x7F; // Takes care of overflows.
    block = permute(block, IP, 64);
    l = block >> 32;
    r = block;
    for(i = 0; i < rounds; i++)
    {
        copy = r;
        r = l ^ f(r, subkeys[i]);
        l = copy;
    }
    block = permute(((uint64_t)r << 32) | l, FP, 64);
    
    return block;
}

int main(argc, argv)
int argc;
char **argv;
{
    uint64_t key, block, temp;
    uint8_t i;
    uint64_t *subkeys;
    
    key = 0x9649f3e19e66266a;
    subkeys = NULL;
    block = 0x68036386d860cba4;
    
    subkeys = des_ksa(key, DES_ROUNDS);
    if(!subkeys)
    {
        printf("error: failed to generate the appropriate subkeys");
        return 1;
    }
    
    temp = des_crypt(subkeys, block, DES_ROUNDS);
    for(i = 0; i < DES_ROUNDS >> 1; i++)
        swap(&subkeys[i], &subkeys[DES_ROUNDS - 1 - i], sizeof(uint64_t));
    printf("%lu vs. %lu\n", block, des_crypt(subkeys, temp, DES_ROUNDS));
    free(subkeys);
    
    return 0;
}

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GCC-6.3.0中的奇怪行为

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