我正在尝试编写SHA-256哈希函数进行练习。在wikipedia表示初始散列值由前8个素数2..19的平方根的小数部分给出。现在我正在计算它们。什么到目前为止我已经做:SHA-256的双精度小数部分的十六进制表示
#include <vector>
#include <cstdint>
#include <cmath>
#include <cstdio>
// fill primes with all prime values between min and max value
int getPrimes(uint32_t min, uint32_t max, std::vector<uint32_t>* primes)
{
if (min < 1) min = 1; // primes can only be >= 1
if (min > max) return 0; // max has to be larger than min
for (uint32_t value = min; value <= max; value++)
{
uint32_t tmp;
for (tmp = 2; tmp <= sqrt(value); tmp++) // start to check with 2, because 1 is always going to work
{
if (value % tmp == 0)
{
break;
}
}
if (tmp > sqrt(value)) primes->push_back(value); // if no other integer divisor is found, add number to vector
}
return 0;
}
int main()
{
std::vector<uint32_t> primes;
getPrimes(2, 20, &primes); // fills vector with all prime values between 2 and 20
double tmp = sqrt(primes[0]); // get square root, returns double
printf("value %f\n", tmp); // debug
printf("size of double %i\n", sizeof(double)); // get representation byte size
double * tmpOffset = &tmp; // get value offset
unsigned char * tmpChar = (unsigned char*)tmpOffset; // convert to char pointer
printf("address of variable %i\n", &tmp); // debug
printf("raw values\n1:%X\n2:%X\n3:%X\n4:%X\n5:%X\n6:%X\n7:%X\n8:%X\n",
(uint8_t)tmpChar[0], (uint8_t)tmpChar[1], (uint8_t)tmpChar[2], (uint8_t)tmpChar[3],
(uint8_t)tmpChar[4], (uint8_t)tmpChar[5], (uint8_t)tmpChar[6], (uint8_t)tmpChar[7]);
return 0;
}
这将返回第一8个素数,计算出的2的平方根,并直接从存储它的实际字节值的存储器位置读取:
value 1.414214
size of double 8
address of variable 6881016
raw values
1:CD
2:3B
3:7F
4:66
5:9E
6:A0
7:F6
8:3F
与维基百科文章0x6a09e667
中给出的值相比,这看起来非常错误,我在这里做的是什么。是否有重新映射发生,或者双重重新表达是多么令人兴奋?有人能指出我如何正确计算十六进制中的小数部分吗?
编辑: 感谢您的帮助!它不漂亮,但现在工作。
printf("raw fractional part:\n0x%02X %02X %02X %02X %02X %02X %02X\n",
(uint8_t)(0xf & tmpChar[6]), (uint8_t)tmpChar[5], (uint8_t)tmpChar[4], (uint8_t)tmpChar[3],
(uint8_t)tmpChar[2], (uint8_t)tmpChar[1], (uint8_t)tmpChar[0]);
uint32_t fracPart = (0xf & tmpChar[6]);
fracPart <<= 8;
fracPart |= tmpChar[5];
fracPart <<= 8;
fracPart |= tmpChar[4] ;
fracPart <<= 8;
fracPart |= tmpChar[3];
fracPart <<= 4;
fracPart |= (0xf0 & tmpChar[2]) >> 4;
printf("fractional part: %X\n", fracPart);
EDIT2 一个更好的实现一点点:
uint32_t fracPart2 = *(uint32_t*)((char*)&tmp + 3); // point to fractional part - 4 bit
fracPart2 <<= 4; // shift to correct value
fracPart2 |= (0xf0 & *((char*)&tmp + 2)) >> 4; // append last 4 bit
printf("beautiful fractional part: %X\n", fracPart2);
该解决方案具有高度的平台依赖性,并在第二个方法我打算像在评论2的链接。
EDIT3
所以这是我最终的解决方案,它不依赖于double的内部表示,并只使用数学计算分数。
uint32_t getFractionalPart(double value)
{
uint32_t retValue = 0;
for (uint8_t i = 0; i < 8; i++)
{
value = value - floor(value);
retValue <<= 4;
value *= 16;
retValue += floor(value);
}
return retValue;
}
当你怀疑,双打的二进制表示和漂浮不是那么简单。我建议在数学上创建十六进制字符串,而不是乱搞字节和可疑的指针转换。 (反复乘以16得到整数部分) – qxz
0x6a09e667表示√2是1.6a09e667 ...¹六进制,读取http://stackoverflow.com/questions/20650954/how-to-convert-decimal-fractions-to-十六进制分数。 'double'不是这样表示的。阅读https://en.wikipedia.org/wiki/Double-precision_floating-point_format。 – kennytm