Why Can't I Use `if constexpr` in Non-Templated Functions in C 17?

C 17's "if constexpr" Limited to Template Functions

In C 17, the introduction of "if constexpr" has sparked interest in its ability to conditionally execute code based on compile-time constants. However, an issue arises when attempting to use it in non-templated functions.

Consider the following code:

<code class="cpp">#include <iostream>
#include <type_traits>

void print(auto value)
{
  // Compile-time check if value is a pointer
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl;
  else
    std::cout << "Ref to " << value << std::endl;
}

int main()
{
  // Error when printing a non-pointer
  print(100);
}</code>

While this code compiles without errors in templated functions, it fails with a compilation error when placed in a non-templated function like print. The reason lies in the behavior of "if constexpr."

"if constexpr" is designed to allow conditional compilation of template code. In a template function, the compiler deduces the type of the parameter at compile time, enabling the selection of the appropriate code path based on the type. In this case, when the print function is templated, it can identify whether value is a pointer at compile time and execute the corresponding code branch.

However, in a non-templated function, the type of value is known only at runtime, making "if constexpr" ineffective. The compiler treats both branches of the conditional as executable code, leading to the compilation error when attempting to dereference value in the Ptr to branch when value is an integer.

To avoid this issue, it is necessary to place "if constexpr" within a template function or provide explicit type information to the non-templated function to ensure compile-time type deduction.

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