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 constexpr(i == 5) { return true; }
 else { return false; }
}

However, when I compile this with GCC 7 with the following flags: g++-7 -std=c++1z test.cpp -o test I get the following error message:

test.cpp: In function 'constexpr bool test_int(int)':
test.cpp:3:21: error: 'i' is not a constant expression
 if constexpr(i == 5) { return true; }

However, if I replace test_int with a different function:

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

Then the following code compiles with no errors:

int main() {
  constexpr int i = 5;
  static_assert(test_int_no_if(i));
  return 0;
}

I don't understand why the constexpr-if version fails to compile, especially since the static_assert works just fine.

Any advice on this would be appreciated.

Thanks!

why constexpr(i == 5) and not just i==5?

deW1
– deW1

2017-07-28 15:34:59 +00:00
Commented Jul 28, 2017 at 15:34 — deW1
– deW1, Commented Jul 28, 2017 at 15:34
any why so complicated? why not return i==5?

deW1
– deW1

2017-07-28 15:45:53 +00:00
Commented Jul 28, 2017 at 15:45 — deW1
– deW1, Commented Jul 28, 2017 at 15:45

Edgar Rokjān · Accepted Answer · 2017-07-28 15:54:46Z

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.

Chris Uzdavinis · Accepted Answer · 2017-07-28 15:58:46Z

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.)

answered Jul 28, 2017 at 15:58

Chris Uzdavinis

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1 Comment

OznOg Over a year ago

This is the most useful answer IMHO

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