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，该函数用于keccak函数。所以这是函数：
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。 digestInHex中的十六进制字符串保留给定输入的哈希输出。
总的来说，我需要使用切换案例比较新旧项目的时间值。为此，我需要运行keccak 1百万次才能看到更清晰的时差。没有看到哈希输出一百万次，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中进行了十六进制注释的printf的原因。
另外，我也想在切换情况下显示哈希输出。但是当我这样做时它会打印null。因此，如何打印哈希输出，如“ 4d741b6f1eb29cb2a9b9911c82f56fa8d73b04959d3d9d222895df6c0b28aa15”？这是主要的：
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;
}
最佳答案
要获得对分配的内存的引用，并在下面的digestInHex()中将其设置为main()，请传递对指针hex的引用。
在（重新）分配它但在打印之后，还要调整它使其释放。
为此，请如下调整代码：
更改
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);
在
free()末尾删除对free(hex)的呼叫。在
digestInHex()中，将所有digestInHex()更改为hex。更改
void keccak(const char* str, enum bitLength hashValueBitLength, char* hex)
{
成为
void keccak(const char* str, enum bitLength hashValueBitLength, char** hex)
{
仅使用
(*hex)保留呼叫digestInHex()。在
hex中定义和初始化：char * hex = NULL;
将所有呼叫更改为
main()和keccak()以接听digestInHex)但接听&hex。同样，在
hex中添加最后一个main()后再返回。