How Can We Implement Template Selection Based on the Absence of a Function in C Metaprogramming?

Metaprogramming: Template Selection Based on Function Overload

In C metaprogramming, defining template functions conditionally based on type properties is a common technique. However, in some cases, it can be challenging to define the opposite scenario, where a template is selected based on the absence of a particular function.

Specifically, in the example provided, the goal is to define a template function stringify that selects between two implementations:

• If std::to_string is defined for the given type, use to_string(t) to convert the value to a string.
• If std::to_string is not defined for the type, use ostringstream() << t.

The problem arises when trying to express the latter condition. The following unsuccessful attempt attempts to use nested enable_if templates to check if to_string is not defined:

template<typename T> enable_if_t<!decltype(to_string(T{})::value, string> (T t){
    return static_cast<ostringstream&>(ostringstream() << t).str();
}

To solve this issue, we can utilize Walter Brown's void_t type trait:

template <typename...>
using void_t = void;<p>Using this, we can define the has_to_string trait as follows:</p>
<pre class="brush:php;toolbar:false">template<typename T, typename = void>
struct has_to_string
: std::false_type { };

template<typename T>
struct has_to_string<T, 
    void_t<decltype(std::to_string(std::declval<T>()))>
    > 
: std::true_type { };

Now, the stringify template can be defined using conditional template specialization based on the has_to_string trait:

template<typename T>
enable_if_t<has_to_string<T>::value, string> stringify(T t){
    return std::to_string(t);
}

template<typename T>
enable_if_t<!has_to_string<T>::value, string> stringify(T t){
    return static_cast<ostringstream&>(ostringstream() << t).str();
}

This solution effectively selects the appropriate implementation of stringify based on whether std::to_string is defined for the given type.

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