Question

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Is it possible to write a C++ template to check for a function's existence?

say there are 2 classes:

struct A{ int GetInt(){ return 10; } };
struct B{ int m; };

I want to use object of type A or B in following function

tempate< typename T >
int GetInt( const T & t )
{
   //if it's A, I'll call: return t.GetInt();
   //if its' B, I'll call: return t.m;
}

Now, because there are whole bunch of classes, some contain GetInt(), some don’t, I don’t want to write specialization for each type, I only want to distinguish them by ‘containing GetInt() or not in compile time‘, how should I do this ?

Answer 1

Stealing from here, and assuming you fix your code so GetInt is const, we get:

HAS_MEM_FUNC(GetInt, has_GetInt);

template <bool B>
struct bool_type
{
    static const bool value = B;
};

typedef bool_type<true> true_type;
typedef bool_type<false> false_type;

namespace detail
{
    template <typename T>
    int get_int(const T& pX, true_type)
    {
        return pX.GetInt();
    }

    template <typename T>
    int get_int(const T& pX, false_type)
    {
        return pX.m;
    }
}

template <typename T>
int get_int(const T& pX)
{
    return detail::get_int(pX,
                            has_GetInt<T, int (T::*)() const>::value);
}

This is pretty awful design though. You should fix the problem rather than apply a patch.