c - 如何在C中打印另一个函数使用的函数参数？

Question

我需要在主函数中打印一个字符串(名为十六进制)。该字符串是digestInHex函数的参数，该函数在keccak函数中使用。所以这里是函数:

void digestInHex(unsigned long long state[][5], unsigned char *buffer,
                 unsigned char *bufferLocation, int bufferSize, int size, char *hex)
{
    unsigned char *byte;
    int i, j;
    unsigned long long *x;

    const char *hexValues = "0123456789abcdef";

    byte = (unsigned char *)malloc(size * sizeof(char));

    hex = (char *)malloc(((size << 1) + 1) * sizeof(char));

    hex[size << 1] = '\0';

    /* Padding */
    bufferLocation[0] = 1;
    ++bufferLocation;

    while (bufferLocation != &buffer[bufferSize / 8])
    {
        bufferLocation[0] = 0;
        ++bufferLocation;
    }

    buffer[(bufferSize >> 3) - 1] |= 0x80;

    bufferLocation = buffer;

    x = (unsigned long long *)buffer;

    for (j = 0; j * 64 < bufferSize; ++j)
    {
        state[j / 5][j % 5] |= x[j];
    }

    round(state);

    /* Squeezing */
    memcpy(byte, state, size);
    reset(state);

    bufferLocation = buffer;

    for (i = 0; i < size; ++i)
    {
        hex[i << 1] = hexValues[byte[i] >> 4];
        hex[(i << 1) + 1] = hexValues[byte[i] & 15];
    }

    free(byte);

    // printf("%s\n", hex);

    free(hex);
}

void keccak(const char *str, enum bitLength hashValueBitLength, char *hex)
{
    int i = 0;
    int j;
    unsigned char *buffer;
    unsigned char *bufferLocation;
    const int bufferSize = 1600 - (hashValueBitLength * 2);
    unsigned long long *x;
    unsigned long long state[5][5];

    buffer = (unsigned char *)malloc(bufferSize * sizeof(char));
    bufferLocation = buffer;

    reset(state);

    while (str[i] != '\0')
    {
        bufferLocation[0] = (unsigned char)str[i];
        ++bufferLocation;

        if (bufferLocation == &buffer[bufferSize / 8])
        {
            bufferLocation = buffer;
            x = (unsigned long long *)buffer;

            for (j = 0; j * 64 < bufferSize; ++j)
            {
                state[j / 5][j % 5] |= x[j];
            }

            round(state);
        }

        ++i;
    }

    digestInHex(state, buffer, bufferLocation, bufferSize, hashValueBitLength / 8, hex);

    free(buffer);
}

如您所见，keccak 函数最后使用了digestInHex 函数。摘要InHex 中的十六进制字符串保留给定输入的哈希输出。

主要是我需要使用 switch-case 比较我的新旧项目的时间值。为此，我需要运行 keccak 100 万次才能更清楚地看到时间差异。不看哈希输出100万次我无法直接在digestInHex中打印十六进制字符串，这就是为什么我在digestInHex中做了十六进制注释的printf。

此外，我还想在 switch-case 中显示哈希输出。但当我这样做时它会打印 null 。那么如何打印像“4d741b6f1eb29cb2a9b9911c82f56fa8d73b04959d3d9d222895df6c0b28aa15”这样的哈希输出？主要内容如下:

int main()
{
    int i;
    clock_t begin, end;
    double timeSpend;
    int n;

    printf("Enter 1 to see Old Project's time value\n");
    printf("Enter 2 to see New Project's time value\n\n");
    printf("Enter 3 to see Old Project's hash output\n");
    printf("Enter 4 to see New Project's hash output\n\n");
    printf("Please enter a value according to above: ");

iterator:

    scanf_s("%d", &n);

    switch (n)
    {
    case 1:
        begin = clock();
        for (i = 0; i < 1000000; ++i)
            keccakOld("The quick brown fox jumps over the lazy dog", KECCAK_256, hexOld);
        end = clock();
        timeSpend = (double)(end - begin) / CLOCKS_PER_SEC;

        printf("%f sec.\n", timeSpend);

        break;

    case 2:
        begin = clock();
        for (i = 0; i < 1000000; ++i)
            keccak("The quick brown fox jumps over the lazy dog", KECCAK_256, hex);
        end = clock();
        timeSpend = (double)(end - begin) / CLOCKS_PER_SEC;

        printf("%f sec.\n", timeSpend);

        break;

    case 3:
        keccakOld("The quick brown fox jumps over the lazy dog", KECCAK_256, hexOld);
        printf("%s\n", hexOld);

        break;

    case 4:
        keccak("The quick brown fox jumps over the lazy dog", KECCAK_256, hex);
        printf("%s\n", hex);

        break;

    default:
        printf("Please re-enter a correct value: ");
        goto iterator;

        break;
    }

    return 0;
}

Answer 1

获取对分配的内存的引用并在下面设置 digestInHex()进入main()传递对指针 hex 的引用.

同时调整free()在(重新)分配之前但在打印之后对其进行编译以释放它。

为此，请按如下方式调整代码:

改变

void digestInHex(unsigned long long state[][5], unsigned char* buffer, 
  unsigned char* bufferLocation, int bufferSize, int size, char* hex)
{

成为

void digestInHex(unsigned long long state[][5], unsigned char* buffer, 
  unsigned char* bufferLocation, int bufferSize, int size, char ** hex)
{
  free(*hex);

删除对 free(hex) 的调用在 digestInHex() 末尾.

在 digestInHex()全部更改hex至(*hex) .

改变

void keccak(const char* str, enum bitLength hashValueBitLength, char* hex)
{

成为

void keccak(const char* str, enum bitLength hashValueBitLength, char** hex)
{

留下电话digestInHex()因为它只使用 hex .

在 main()定义和初始化:

char * hex = NULL;

将所有调用更改为 keccak()和digestInHex)采取&hex但是hex .

也在 main()添加最后一个 free(hex)返回之前。