check if member exists using enable_if

Question

Here\'s what I\'m trying to do:

template  struct Model
{
    vector vertices ;

    #if T has a .normal member
    void transform(         


        

Answer 1

I know this question already has some answers but I think my solution to this problem is a bit different and could help someone.

The following example checks whether passed type contains c_str() function member:

template 
struct has_c_str : false_type {};

template 
struct has_c_str> : std::is_same().c_str())>
{};

template ::value, StringType>::type* = nullptr>
bool setByString(StringType const& value) {
    // use value.c_str()
}

In case there is a need to perform checks whether passed type contains specific data member, following can be used:

template 
struct has_field : std::false_type {};

template 
struct has_field> : std::is_convertible
{};

template ::value, T>::type* = nullptr>
void fun(T const& value) {
    // use value.field ...
}

UPDATE C++20

C++20 introduced constraints and concepts, core language features in this C++ version.

If we want to check whether template parameter contains c_str member function, then, the following will do the work:

template
concept HasCStr = requires(T t) { t.c_str(); };

template  
void setByString(StringType const& value) {
    // use value.c_str()
}

Furthermore, if we want to check if the data member, which is convertible to long, exists, following can be used:

template
concept HasField = requires(T t) {
    { t.field } -> std::convertible_to;
};

template  
void fun(T const& value) {
    // use value.field
}

By using C++20, we get much shorter and much more readable code that clearly expresses it's functionality.