Swap bits in c++ for a double

Question

Im trying to change from big endian to little endian on a double. One way to go is to use

double val, tmp = 5.55;

((unsigned int *)&val)[0] = ntohl(((unsigned int *)&tmp)[1]);
((unsigned int *)&val)[1] = ntohl(((unsigned int *)&tmp)[0]);

But then I get a warning: "dereferencing type-punned pointer will break strict-aliasing rules" and I dont want to turn this warning off.

Another way to go is:

#define ntohll(x) ( ( (uint64_t)(ntohl( (uint32_t)((x << 32) >> 32) )) << 32) | ntohl( ((uint32_t)(x >> 32)) ) ) 

val = (double)bswap_64(unsigned long long(tmp)); //or
val = (double)ntohll(unsigned long long(tmp));

But then a lose the decimals. Anyone know a good way to swap the bits on a double without using a for loop?

Answer 1

I'd probably try something like this:

template <typename T>
void swap_endian(T& pX)
{
    // should static assert that T is a POD

    char& raw = reinterpret_cast<char&>(pX);
    std::reverse(&raw, &raw + sizeof(T));
}

Short and sweet (and relatively untested). The compiler will make all the necessary optimizations. The above is well-defined for any POD type, and doesn't rely on any implementation details.

A copy version, for when you don't want to modify the argument:

template <typename T>
T swap_endian_copy(T pX)
{
    swap_endian(pX);
    return pX;
}

Answer 2

There are some important pitfalls to pay attention to when dealing with the binary representation of floats or doubles.