Comparing constexpr function parameter in constexpr-if condition causes error

Question

I\'m trying to compare a function parameter inside a constexpr-if statement.

Here is a simple example:

constexpr bool test_int(const int i) {
  if c         


        

Answer 1

From constexpr if:

In a constexpr if statement, the value of condition must be a contextually converted constant expression of type bool.

Then, from constant expression:

Defines an expression that can be evaluated at compile time.

Obviously, i == 5 is not a constant expression, because i is a function parameter which is evaluated at run time. That is why the compiler complains.

When you use a function:

constexpr bool test_int_no_if(const int i) { return (i == 5); }

then it might be evaluated during the compile time depending on whether it's parameter is known at compile time or not.

If i is defined like:

constexpr int i = 5;

then the value of i is known during the compile time and test_int_no_if might be evaluated during the compile too making it possible to call it inside static_assert.

Also note, that marking function parameter as const does not make it a compile time constant. It just means that you cannot change the parameter inside the function.

Answer 2

A constexpr function can be called with non-constexpr arguments, in which case it behaves like a normal function, so the code must still compile as if it were not constexpr.

In short, there's nothing in test_int_no_if that depends on i being constexpr, while in test_int(), there is. ("constexpr if" only works with compile time expressions.)