Converting a double to intptr_t in C++

Question

Assume the following:

#include <iostream>

int main()
{
   double a = 5.6;
   intptr_t b = (intptr_t) a;
   double c = (double) b;
   return 0;
}

c will be 5 . My question is, since intptr_t is also 64 bits on a 64 bit machine (same as double), how come the precision bits are not saved during casting?

Answer 1

Although intptr_t is meant to represent a pointer to an int, its underling type is still an integer. Thus

intptr_t b = (intptr_t)a

Is still truncating the double , similar to if you'd just written:

int b = (int)a;

What you want to do is take the address of a :

intptr_t b = (intptr_t)&a

And then convert it back

double c = *(double*)b;