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;
来源：https://stackoverflow.com/questions/17409162/double-from-long-bits