Double from long bits

问题:

I have an unsigned long long (or uint64_t) value and want to convert it to a double. The double shall have the same bit pattern as the long value. This way I can set the bits of the double "by hand".

unsigned long long bits = 1ULL;
double result = /* some magic here */ bits;

I am looking for a way to do this.

回答1:

The portable way to do this is with memcpy (you may also be able to conditionally do it with reinterpret_cast or a union, but those aren't certain to be portable because they violate the letter of the strict-alias rules):

// First, static assert that the sizes are the same
memcpy(&result, &bits, sizeof(bits));

But before you do make sure you know exactly what you're doing and what floating point representation is being used (although IEEE754 is a popular/common choice). You'll want to avoid all kinds of problem values like infinity, NaN, and denormal numbers.

回答2:

Beware of union and reinterpret_cast<double*>(&bits), for both of these methods are UB. Pretty much all you can do is memcpy.

回答3:

The following uses a void pointer.

unsigned long long bits = 1ULL;
void* tempPtr=(void*)&bits;
double result = *(double*)tempPtr;