Howto check a type for the existence of parameterless operator()

Question

I\'m trying to check whether a functor is compatible with a given set of parametertypes and a given return type (that is, the given parametertypes can be implicitely converted t

Answer 1

Have you tried something like:

template
class Discrim
{
};

template
std::true_type hasFunctionCallOper( T*, Discrim* );

template
std::false_type hasFunctionCallOper( T*, ... );

After, you discriminate on the return type of hasFunctionCallOper((T*)0, 0).

EDITED (thanks to the suggestion of R. Martinho Fernandes):

Here's the code that works:

template
class CallOpDiscrim {};

template
TrueType hasCallOp( T*, CallOpDiscrim< sizeof( (*((T const*)0))(), 1 ) > const* );
template
FalseType hasCallOp( T* ... );

template
class TestImpl;

template
class TestImpl
{
public:
    void doTellIt() { std::cout << typeid(T).name() << " does not have operator()" << std::endl; }
};

template
class TestImpl
{
public:
    void doTellIt() { std::cout << typeid(T).name() << " has operator()" << std::endl; }
};

template
class Test : private TestImpl(0, 0)) == sizeof(TrueType)>
{
public:
    void tellIt() { this->doTellIt(); }
};