Execute function inside function template only for those types that have the function defined

Question

I have a function template which takes many different types as it's input. Out of those types only one of them has a getInt() function. Hence I want the code to run the function only for that type. Please suggest a solution. Thanks

#include <type_traits>
#include <typeinfo>

class X {
    public:
    int getInt(){
        return 9;
    }
};

class Y{

};

template<typename T>
void f(T& v){
    // error: 'class Y' has no member named 'getInt'
    // also tried std::is_same<T, X>::value 
    if(typeid(T).name() == typeid(X).name()){
        int i = v.getInt();// I want this to be called for X only
    }
}

int main(){
    Y y;
    f(y);
}

Answer 1

If you want to be able to call a function f for all types that have function member getInt, not just X, you can declare 2 overloads for function f:

1. for types that have getInt member function, including class X

2. for all the other types, including class Y.

C++11 / C++17 solution

Having that in mind, you could do something like this:

#include <iostream>
#include <type_traits>

template <typename, typename = void>
struct has_getInt : std::false_type {};

template <typename T>
struct has_getInt<T, std::void_t<decltype(((T*)nullptr)->getInt())>> : std::is_convertible<decltype(((T*)nullptr)->getInt()), int>
{};

class X {
public:
    int getInt(){
        return 9;
    }
};

class Y {};

template <typename T,
          typename std::enable_if<!has_getInt<T>::value, T>::type* = nullptr>
void f(T& v) {
    // only for Y
    std::cout << "Y" << std::endl;
}

template <typename T,
          typename std::enable_if<has_getInt<T>::value, T>::type* = nullptr>
void f(T& v){
    // only for X
    int i = v.getInt();
    std::cout << "X" << std::endl;
}

int main() {
    X x;
    f(x);

    Y y;
    f(y);
}

Check it out live.

Please note that std::void_t is introduced in C++17, but if you are limited to C++11, then it is really easy to implement void_t on your own:

template <typename...>
using void_t = void;

And here is C++11 version live.

What do we have in C++20?

C++20 brings lots of good things and one of them is concepts. Above thing that's valid for C++11/C++14/C++17 can be significantly reduced in C++20:

#include <iostream>
#include <concepts>

template<typename T>
concept HasGetInt = requires (T& v) { { v.getInt() } -> std::convertible_to<int>; };

class X {
public:
    int getInt(){
        return 9;
    }
};

class Y {};

template <typename T>
void f(T& v) {
    // only for Y
    std::cout << "Y" << std::endl;
}

template <HasGetInt T>
void f(T& v){
    // only for X
    int i = v.getInt();
    std::cout << "X" << std::endl;
}

int main() {
    X x;
    f(x);

    Y y;
    f(y);
}

Check it out live.

Answer 2

You might use if constexpr from C++17:

template<typename T>
void f(T& v){
    if constexpr(std::is_same_v<T, X>) { // Or better create trait has_getInt
        int i = v.getInt();// I want this to be called for X only
    }
    // ...
}

Before, you will have to use overloads and SFINAE or tag dispatching.

Answer 3

Keep it simple and overload. Has worked since at least C++98...

template<typename T>
void f(T& v)
{
    // do whatever
}

void f(X& v)
{
    int result = v.getInt();
}

This is enough if there only ever one type with getInt function. If there's more, it's not so simple anymore. There are several ways to do it, here's one:

struct PriorityA { };
struct PriorityB : PriorityA { };

template<typename T>
void f_impl(T& t, PriorityA)
{
    // generic version
}

// use expression SFINAE (-> decltype part)
// to enable/disable this overload
template<typename T>
auto f_impl(T& t, PriorityB) -> decltype(t.getInt(), void())
{
    t.getInt();
}

template<typename T>
void f(T& t)
{
    f_impl(t, PriorityB{ } ); // this will select PriorityB overload if it exists in overload set
                              // otherwise PriorityB gets sliced to PriorityA and calls generic version
}

Live example with diagnostic output.