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>)
    std::cout <p>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.</p>
<p><strong>Why Only Templates?</strong></p>
<p>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<decltype>. However, if the condition is false, the compiler still attempts to dereference the integer 'value', which is invalid.</decltype></p>
<p><strong>Solution</strong></p>
<p>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:</p>
<pre class="brush:php;toolbar:false"><code class="cpp">#include <iostream>
#include <type_traits>

template<typename t>
void print(T value)
{
  if constexpr (std::is_pointer_v<t>)
    std::cout <p>In this case, the print function is templated, so if constexpr behavior is correctly applied based on the template parameter.</p></t></typename></type_traits></iostream></code>

The above is the detailed content of Why Doesn't `if constexpr` Work in Non-Templated Functions in C 17?. For more information, please follow other related articles on the PHP Chinese website!