问题:

I've got a byte array containing 8 bytes and would like to convert and use them as a double precision binary floating-point number.

Could someone please tell me how to convert it?

回答1:

Try this:

double a;
memcpy(&a, ptr, sizeof(double));

where ptr is the pointer to your byte array. If you want to avoid copying use a union, e.g.

union {
  double d;
  char bytes[sizeof(double)];
} u;
// Store your data in u.bytes
// Use floating point number from u.d

回答2:

Here is one solution using memcpy function:

double d = 0;
unsigned char buf[sizeof d] = {0};

memcpy(&d, buf, sizeof d);

Note that a solution like:

d = *((double *) buf);

shoud be avoided. This is undefined behavior because it potentially violates alignment and aliasing rules.

回答3:

In C++:

double x;
char buf[sizeof(double)]; // your data

#include <algorithm>

// ...
std::copy(buf, buf + sizeof(double), reinterpret_cast<char*>(&x));

In C:

#include <string.h>

/* ... */
memcpy(&x, buf, sizeof(double));

In C++11, you can also use std::begin(buf) and std::end(buf) as the boundaries (include the header <iterator>), and in both languages you can use sizeof(buf) / sizeof(buf[0]) (or simply sizeof(buf)) for the size, all provided buf is actually an array and not just a pointer.