Question

Consider the following piece of code.

int var;

cout << (long)&var;

My doubt is how do we know that long int has sufficient width to hold the memory location indicated by &var. What if it is not sufficient?

The full code which I am executing…

//: C03:YourPets2.cpp
// From Thinking in C++, 2nd Edition
// Available at http://www.BruceEckel.com
// (c) Bruce Eckel 2000
// Copyright notice in Copyright.txt
#include <iostream>
using namespace std;

int dog, cat, bird, fish;

void f(int pet) {
  cout << "pet id number: " << pet << endl;
}

int main() {
  int i, j, k;
  cout << "Address size " << sizeof(&f) << endl;
  cout << "Long size " << sizeof(long) << endl;
  cout << "Intptr size " << sizeof(intptr_t) << endl;
  cout << "f(): " << &f << endl;  
  cout << "f(): " << (long)&f << endl;
  cout << "f(): " << (long long)&f << endl;
  cout << "dog: " << (long)&dog << endl;
  cout << "cat: " << &cat << endl;
  cout << "bird: " << &bird << endl;
  cout << "fish: " << (long)&fish << endl;
  cout << "i: " << (long)&i << endl;
  cout << "i: " << (long long)&i << endl;
  cout << "j: " << (long)&j << endl;
  cout << "k: " << (long)&k << endl;
} ///:~

The result which I am getting:

Address size 4
Long size 4
Intptr size 4
f(): 1
f(): 134514548
f(): 134514548
dog: 134521044
cat: 0x804a0d8
bird: 0x804a0dc
fish: 134521056
i: -1074729380
i: -1074729380
j: -1074729384
k: -1074729388

Answer 1

You don’t. It’s possible – if unlikely – for pointers to have larger storage requirements than any integer. If there is an integer type that is suitable then there will be a typedef for it std::intptr_t (and possible also std::uintptr_t) defined in <cstdint> (C++11 only).

You can test for the presence of intptr_t at the preprocessor stage by testing for the definedness of the macro INTPTR_MAX (or INTPTR_MIN) after #include <cstdint>.

If you just want to print a pointer value using std::cout then you can cast to void* (unnecessary for int* but necessary for char*) and use << directly without a cast to an integer type.