Perfect forward a const parameter for constant evaluation

Question

I am very confused by why the following code doesn't compile:

template <typename T, typename... Ts> 
void TEST(T&& t, Ts&&... ts) {
  if constexpr(sizeof...(ts) > 0) {
    TEST(std::forward<Ts>(ts)...);
    static_assert( t == 2 , "parameter must be 2");
  }
}

template <typename... Ts> 
void VARIADIC(Ts&&... ts) {
  TEST(std::forward<Ts>(ts)...);
}

int main(int argc, char* argv[]) {
  constexpr int a1 = 2;
  constexpr int a2 = 2;
  VARIADIC(a1, a2);
}

Basically, I am trying to do some compile-time check on the parameters passed to the function VARIADIC. However, the compiler complained about the following:

error: non-constant condition for static assertion
     static_assert( t == 2 , "parameter must be 2");
     ^~~~~~~~~~~~~
error: ‘t’ is not a constant expression

It is very obvious the given parameters a1 and a2 are constants, and there must be some ways to perform the check on variadic arguments.

What is the confusion about? Entities are only treated as constant expression when they are introduced as part of a constant expression. Clearly, the parameters to a function are *not* constant expression in that sense. However, the function result *can* be a constant expression. — Dietmar Kühl, May 07 '17 at 22:37
The error is in the definition of the TEST function, which could be called with non-constant values. The fact that you happen to be calling TEST with constant values doesn't matter. — Vaughn Cato, May 07 '17 at 22:37
You could try the following: http://stackoverflow.com/questions/8626055/c11-static-assert-within-constexpr-function — jwimberley, May 08 '17 at 01:53

score 4 · Answer 1 · answered May 07 '17 at 22:45

The function arguments are not core constant expressions according to 5.20 [expr.const] paragraph 2.9: variables or references are only core constant expressions if they are initialized with a constant expression or their live-time began within the definition of constant expression. Function arguments clearly satisfy neither.

The background is that constexpr functions can be called with run-time values. For the determination of whether a variable is a constant expression within the function it is immaterial whether it is actually called with a constant expression. Whether it matter whether the call is with a constant expression is when the function is evaluated: although the arguments do not behave as constant expressions within the function the result can still be a constant expression.

Perfect forward a const parameter for constant evaluation

1 Answers1