c++: constexpr function doesn't evaluate at compile time when using with std::cout [duplicate]

Question

I'm new with c++ and currently came across with constexpr. As I realize constexpr functions are evaluated at compile time. Here is my source code:

constexpr int sum(float a, int b)
{
    return a + b;
};

int main(int argc, char *argv[])
{
    std::cout << sum(1, 2) << std::endl;
}

It is a simple function which just sums to integers. The problem is that when I set breakpoint at return a + b and start debugging, the breakpoint is hit, which means that the function was not evaluated at compile time. But when I change main function to this:

int main(int argc, char *argv[])
{
    constexpr int var = sum(2, 2);
    std::cout << var << std::endl;
}

the breakpoint is not hit, which means that function was evaluated at compile time. I'm little confused why function is not evaluated in first case?

P.S I'm using visual studio 2017.

Answer 1

As I realize constexpr functions are evaluated at compile time

Not really. They can be evaluated at compile-time, but they are not guaranteed to do so, unless they're invoked in a context where a constant expression is required.

One such context is the declaration of a constexpr variable.

Answer 2

constexpr means "can be evaluated at compile time" rather than "must be evaluated at compile time". If you want to see it evaluated at compile time you can call it in a context that requires to be evaluated at compile time, for example a template parameter:

std::array<int, sum(3,5)> x;

Note that the motivation for constexpr is the other way around than many would expect. constexpr tells the compiler that you can use it eg as template parameter and if sum was not constexpr you'd get a compiler error. It is not to make sure that the function always is evaluated at compile time.