How can two versions of the same function, differing only in one being inline and the other one not, return different values? Here is some code I wrote today and I am not sure how it works.

#include <cmath>
#include <iostream>

bool is_cube(double r)
{
    return floor(cbrt(r)) == cbrt(r);
}

bool inline is_cube_inline(double r)
{
    return floor(cbrt(r)) == cbrt(r);
}

int main()
{
    std::cout << (floor(cbrt(27.0)) == cbrt(27.0)) << std::endl;
    std::cout << (is_cube(27.0)) << std::endl;
    std::cout << (is_cube_inline(27.0)) << std::endl;
}

I would expect all outputs to be equal to 1, but it actually outputs this (g++ 8.3.1, no flags):

1
0
1

instead of

1
1
1

Edit: clang++ 7.0.0 outputs this:

0
0
0

and g++ -Ofast this:

1
1
1

ANSWER

Some compilers (notably GCC) use higher precision when evaluating expressions at compile time. If an expression depends only on constant inputs and literals, it may be evaluated at compile time even if the expression is not assigned to a constexpr variable. Whether or not this occurs depends on:

  • The complexity of the expression
  • The threshold the compiler uses as a cutoff when attempting to perform compile time evaluation
  • Other heuristics used in special cases (such as when clang elides loops)

If an expression is explicitly provided, as in the first case, it has lower complexity and the compiler is likely to evaluate it at compile time.

Similarly, if a function is marked inline, the compiler is more likely to evaluate it at compile time because inline functions raise the threshold at which evaluation can occur.

Higher optimization levels also increase this threshold, as in the -Ofast example, where all expressions evaluate to true on gcc due to higher precision compile-time evaluation.

We can observe this behavior here on compiler explorer. When compiled with -O1, only the function marked inline is evaluated at compile-time, but at -O3 both functions are evaluated at compile-time.

NB: In the compiler-explorer examples, I use printf instead iostream because it reduces the complexity of the main function, making the effect more visible.

Demonstrating that inline doesn’t affect runtime evaluation

We can ensure that none of the expressions are evaluated at compile time by obtaining value from standard input, and when we do this, all 3 expressions return false as demonstrated here: https://ideone.com/QZbv6X

#include <cmath>
#include <iostream>

bool is_cube(double r)
{
    return floor(cbrt(r)) == cbrt(r);
}
 
bool inline is_cube_inline(double r)
{
    return floor(cbrt(r)) == cbrt(r);
}

int main()
{
    double value;
    std::cin >> value;
    std::cout << (floor(cbrt(value)) == cbrt(value)) << std::endl; // false
    std::cout << (is_cube(value)) << std::endl; // false
    std::cout << (is_cube_inline(value)) << std::endl; // false
}

Contrast with this example, where we use the same compiler settings but provide the value at compile-time, resulting in the higher-precision compile-time evaluation.