Why Does "if constexpr" Cause Compilation Errors in Non-Templated Functions in C 17?

Error Using "If constexpr" in Non-Templated Functions in C 17

In C 17, the "if constexpr" statement allows for conditionally evaluating and compiling code based on constant expressions. However, this feature has a limitation when used in non-templated functions.

Code:

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

void print(auto value)
{
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl;
  else
    std::cout << "Ref to " << value << std::endl;
}</code>

Problem:

When "if constexpr" is used in a non-templated function, as shown above, a compilation error occurs when attempting to dereference a non-pointer value. This error arises because, unlike in templated functions, "if constexpr" in non-templated functions does not prevent the compilation of code paths that are not taken.

Explanation:

In templated functions, "if constexpr" allows code to be conditionally compiled based on the value of template parameters. In this case, the compiler can determine at compile time whether the template is valid for a given specialization and only instantiate the necessary code path.

However, in non-templated functions, "if constexpr" acts only as a conditional statement at runtime. The compiler still analyzes both branches of the conditional statement, and if any code in the non-taken branch is invalid, a compilation error will occur.

Solution:

To avoid this error, use "if constexpr" only in templated functions or in non-templated functions where the type is known at compile time and is guaranteed to be valid for both branches of the conditional statement.

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