c++ - Comparing constexpr function parameter in constexpr-if condition causes error -

i'm trying compare function parameter inside constexpr-if statement.

here simple example:

constexpr bool test_int(const int i) {   if constexpr(i == 5) { return true; }  else { return false; } }

however, when compile gcc 7 following flags: g++-7 -std=c++1z test.cpp -o test following error message:

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

however, if replace test_int different function:

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

then following code compiles no errors:

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

i don't understand why constexpr-if version fails compile, since static_assert works fine.

any advice on appreciated.

thanks!

in constexpr if statement, value of condition must contextually converted constant expression of type bool.

defines expression can evaluated @ compile time.

obviously, i == 5 not constant expression, because i function parameter evaluated @ run time. why compiler complains.

when use function:

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

then might evaluated during compile time depending on whether it's parameter known @ compile time or not.

if i defined like:

constexpr int = 5;

then value of i known during compile time , test_int_no_if might evaluated during compile making possible call inside static_assert.

also note, marking function parameter const not make compile time constant. means cannot change parameter inside function.

Insert