constexpr question, why do these two different programs run in such a different amount of time with g++?

Question

I'm using gcc 4.6.1 and am getting some interesting behavior involving calling a constexpr function. This program runs just fine and straight away prints out 12200160415121876738.

#include <iostream>

extern const unsigned long joe;

constexpr unsigned long fib(unsigned long int x)
{
   return (x <= 1) ? 1 : (fib(x - 1) + fib(x - 2));
}

const unsigned long joe = fib(92);

int main()
{
   ::std::cout << "Here I am!\n";
   ::std::cout << joe << '\n';
   return 0;
}

This program takes forever to run and I've never had the patience to wait for it to print out a value:

#include <iostream>

constexpr unsigned long fib(unsigned long int x)
{
   return (x <= 1) ? 1 : (fib(x - 1) + fib(x - 2));
}

int main()
{
   ::std::cout << "Here I am!\n";
   ::std::cout << fib(92) << '\n';
   return 0;
}

Why is there such a huge difference? Am I doing something wrong in the second program?

Edit: I'm compiling this with g++ -std=c++0x -O3 on a 64-bit platform.

Answer 1

I also wanted to see how gcc did optimize the code for this new constexpr keyword, and actually it's just because you are calling fib(92) as parameter of the ofstream::operator<<

::std::cout << fib(92) << '\n';

that it isn't evaluated at the compilation time, if you try calling it not as a parameter of another function (like you did in)

const unsigned long joe = fib(92);

it is evaluated at compile time, I did a blog post about this if you want more info, I don't know if this should be mentioned to gcc developers.

Answer 2

joe is an Integral Constant Expression; it must be usable in array bounds. For that reason, a reasonable compiler will evaluate it at compile time.

In your second program, even though the compiler could calculate it at compile time, there's no reason why it must.