如何在 C++ 中将无符号字符缓冲区转换为双精度值？

Question

I am trying to convert an unsigned char buffer array into a double. Why does this not copy the bytes into the double as they are in the array?

#include <iostream>
#include <string>

int main() {

    unsigned char buffer[8] = {63, 240, 0, 0, 0, 0, 0, 0};
    double x = *(double*)buffer;
    std::cout << x << std::endl;
    return 0;
}

I also tried doing this:

#include <iostream>
#include <string>

int main() {

    unsigned char buffer[8] = {63, 240, 0, 0, 0, 0, 0, 0};
    double x ;
    memcpy(&x, buffer, sizeof(double)); //NOW USING MEMCPY
    std::cout << x << std::endl;
    return 0;
}

I looked at this post here , but it only got the same results. The unsigned chars {63, 240, 0, 0, 0, 0, 0, 0} is the representation of the double number 1 .

It outputs: 3.03865e-319 .

Answer 1

You've got your buffer round the wrong way. It should be:

{0, 0, 0, 0, 0, 0, 240, 63}

(on a little-endian machine using IEEE floating point).

Live demo