模板 function 中的 void 参数

Question

I try to create generic convert function. How to create template function than can have void argument? Something like this:

template<typename T1>
bool convert(T1 arg) {
    return true;
}

template<typename T1 = void>
bool convert(T1 arg) {
    return false;
}

void voidFn() {
    return;
}

bool boolFn() {
    return true;
}

void main() {
   cout << convert(boolFn());  //ok return true
   cout << convert(voidFn());  //compile error no matching function 
}

Answer 1

There's no such thing as a "void argument", in C or C++. This is done using ordinary overloading:

template<typename T1>
bool convert(T1 arg) {
    return true;
}

bool convert() {
    return false;
}

Also, depending on the actual details, there's a small chance that it might be possible to have a single template function that uses std::optional and user-defined deduction guides in C++17, and later.

Answer 2

You can't have a void argument for a function.

But you can have void type template parameter. So you need a function that select void /non void returning functions, you can try something as follows

#include <iostream>

template <typename>
bool convert ()
 { return true; }

template <>
bool convert<void> ()
 { return false; }

void voidFn ()
 { return; }

bool boolFn ()
 { return true; }

int main ()
 {
   std::cout << convert<decltype(boolFn())>() << '\n'; // print 1
   std::cout << convert<decltype(voidFn())>() << '\n'; // print 0
 }

Obviously this can be useful only if it's relevant the type, not the exact value, returned from the functions.