Why Doesn't `if constexpr` Work in Non-Templated Functions in C 17?

if constexpr Does Not Behave As Expected in Non-Templated Functions in C 17

In C 17, the if constexpr statement allows for conditional compilation based on compile-time constant expressions. However, when attempting to use this feature in a non-templated function, unexpected compilation errors may occur.

Consider the following code snippet:

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

int main()
{
  auto value = 100;
  if constexpr (std::is_pointer_v<decltype(value)>)
    std::cout << "Ptr to " << *value << std::endl; // Error
  else
    std::cout << "Ref to " << value << std::endl;
}</code>

When compiling this code, an error will occur on the line attempting to dereference 'value' within the if constexpr statement. This is because if constexpr is not supported in non-templated functions.

Why Only Templates?

This behavior is by design. if constexpr is intended to prevent instantiation of invalid template code based on specific specializations. In the example provided, the non-templated function checks whether 'value' is a pointer using std::is_pointer_v. However, if the condition is false, the compiler still attempts to dereference the integer 'value', which is invalid.

Solution

To use if constexpr in non-templated contexts, consider wrapping the code in a template function or using a macro that will behave differently at compile time. For example:

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

template<typename T>
void print(T value)
{
  if constexpr (std::is_pointer_v<T>)
    std::cout << "Ptr to " << *value << std::endl;
  else
    std::cout << "Ref to " << value << std::endl;
}

int main()
{
  auto n = 1000;
  print(n);

  int *ptr = &n;
  print(ptr);
}</code>

In this case, the print function is templated, so if constexpr behavior is correctly applied based on the template parameter.

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