How to Check If a Member Name (Variable or Function) Exists in a Class, with or Without Specifying Type

How to check if a member name (variable or function) exists in a class, with or without specifying type?

C++03

#define HasMember(NAME) \
  template<class Class, typename Type = void> \
  struct HasMember_##NAME \
  { \
    typedef char (&yes)[2]; \
    template<unsigned long> struct exists; \
    template<typename V> static yes Check (exists<sizeof(static_cast<Type>(&V::NAME))>*); \
    template<typename> static char Check (...); \
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
  }; \
  template<class Class> \
  struct HasMember_##NAME<Class, void> \
  { \
    typedef char (&yes)[2]; \
    template<unsigned long> struct exists; \
    template<typename V> static yes Check (exists<sizeof(&V::NAME)>*); \
    template<typename> static char Check (...); \
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
  }

Usage: Simply invoke the macro with whatever member you want to find:

HasMember(Foo);  // Creates a SFINAE `class HasMember_Foo`
HasMember(i);    // Creates a SFINAE `class HasMember_i`

Now we can utilize HasMember_X to check X in ANY class as below:

#include<iostream>
struct S
{
  void Foo () const {}
//  void Foo () {}  // If uncommented then type should be mentioned in `HasMember_Foo`    
  int i;
};
int main ()
{
  std::cout << HasMember_Foo<S, void (S::*) () const>::value << "\n";
  std::cout << HasMember_Foo<S>::value << "\n";
  std::cout << HasMember_i<S, int (S::*)>::value << "\n";
  std::cout << HasMember_i<S>::value << "\n";
}

Catches:

1. In case of methods, if we don't mention the type then the class
   must not have overloaded methods. If it has then this trick fails.
   i.e. even though the named member is present more than once, the result will be false.
2. If the member is part of base class, then this trick fails; e.g. if B is base of S & void B::Bar () is present, then HasMember_Bar<S, void (B::*)()>::value or HasMember_Bar<S, void (S::*)()>::value or HasMember_Bar<S>::value will give false

Test if a not class member function exists (SFINAE)

You may use something like this:

template <class T>
static auto hasPrintMethod(int)
->std::integral_constant<bool, std::is_class<decltype(print(T()))>::value>;

template <class>
static auto hasPrintMethod(...)->std::false_type;

template <typename T>
struct HasPrintMethod : decltype(hasPrintMethod<T>(0)) {
};

Here the decltype(print(T())) is std::string for classes for which your non-member function is defined, and an erroneous type for other classes. So, according to SFINAE concept, HasPrintMethod<A>::value is equal to true for class A with print function defined and is equal to false otherwise.

Templated check for the existence of a class member function?

Yes, with SFINAE you can check if a given class does provide a certain method. Here's the working code:

#include <iostream>

struct Hello
{
    int helloworld() { return 0; }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_helloworld
{
    typedef char one;
    struct two { char x[2]; };

    template <typename C> static one test( decltype(&C::helloworld) ) ;
    template <typename C> static two test(...);    

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};
    
int main(int argc, char *argv[])
{
    std::cout << has_helloworld<Hello>::value << std::endl;
    std::cout << has_helloworld<Generic>::value << std::endl;
    return 0;
}

I've just tested it with Linux and gcc 4.1/4.3. I don't know if it's portable to other platforms running different compilers.

How to detect whether there is a specific member variable in class?

Another way is this one, which relies on SFINAE for expressions too. If the name lookup results in ambiguity, the compiler will reject the template

template<typename T> struct HasX { 
    struct Fallback { int x; }; // introduce member name "x"
    struct Derived : T, Fallback { };

    template<typename C, C> struct ChT; 

    template<typename C> static char (&f(ChT<int Fallback::*, &C::x>*))[1]; 
    template<typename C> static char (&f(...))[2]; 

    static bool const value = sizeof(f<Derived>(0)) == 2;
}; 

struct A { int x; };
struct B { int X; };

int main() { 
    std::cout << HasX<A>::value << std::endl; // 1
    std::cout << HasX<B>::value << std::endl; // 0
}

It's based on a brilliant idea of someone on usenet.

Note: HasX checks for any data or function member called x, with arbitrary type. The sole purpose of introducing the member name is to have a possible ambiguity for member-name lookup - the type of the member isn't important.

How to check if a function exists in C/C++?

While other replies are helpful advices (dlsym, function pointers, ...), you cannot compile C++ code referring to a function which does not exist. At minimum, the function has to be declared; if it is not, your code won't compile. If nothing (a compilation unit, some object file, some library) defines the function, the linker would complain (unless it is weak, see below).

But you should really explain why you are asking that. I can't guess, and there is some way to achieve your unstated goal.

Notice that dlsym often requires functions without name mangling, i.e. declared as extern "C".

If coding on Linux with GCC, you might also use the weak function attribute in declarations. The linker would then set undefined weak symbols to null.

addenda

If you are getting the function name from some input, you should be aware that only a subset of functions should be callable that way (if you call an arbitrary function without care, it will crash!) and you'll better explicitly construct that subset. You could then use a std::map, or dlsym (with each function in the subset declared extern "C"). Notice that dlopen with a NULL path gives a handle to the main program, which you should link with -rdynamic to have it work correctly.

You really want to call by their name only a suitably defined subset of functions. For instance, you probably don't want to call this way abort, exit, or fork.

NB. If you know dynamically the signature of the called function, you might want to use libffi to call it.

Check if a class has a member function of a given signature

I'm not sure if I understand you correctly, but you may exploit SFINAE to detect function presence at compile-time. Example from my code (tests if class has member function size_t used_memory() const).

template<typename T>
struct HasUsedMemoryMethod
{
    template<typename U, size_t (U::*)() const> struct SFINAE {};
    template<typename U> static char Test(SFINAE<U, &U::used_memory>*);
    template<typename U> static int Test(...);
    static const bool Has = sizeof(Test<T>(0)) == sizeof(char);
};

template<typename TMap>
void ReportMemUsage(const TMap& m, std::true_type)
{
        // We may call used_memory() on m here.
}
template<typename TMap>
void ReportMemUsage(const TMap&, std::false_type)
{
}
template<typename TMap>
void ReportMemUsage(const TMap& m)
{
    ReportMemUsage(m, 
        std::integral_constant<bool, HasUsedMemoryMethod<TMap>::Has>());
}

Python check if function exists without running it

You can use dir to check if a name is in a module:

>>> import os
>>> "walk" in dir(os)
True
>>>

In the sample code above, we test for the os.walk function.

How to check if function exists in JavaScript?

Try something like this:

if (typeof me.onChange !== "undefined") { 
    // safe to use the function
}

or better yet (as per UpTheCreek upvoted comment)

if (typeof me.onChange === "function") { 
    // safe to use the function
}

Type trait: Check if class have specific function (maybe inherit)

Here is one old school C++03 way of doing it. Typically it can be used as a utility and get it molded for any method or variable.

#define HasMember(NAME) \
  template<class Class, typename Type = void> \
  struct HasMember_##NAME \
  { \
    typedef char (&yes)[2]; \
    template<unsigned long> struct exists; \
    template<typename V> static yes Check (exists<sizeof(static_cast<Type>(&V::NAME))>*); \
    template<typename> static char Check (...); \
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
  }; \
  template<class Class> \
  struct HasMember_##NAME<Class, void> \
  { \
    typedef char (&yes)[2]; \
    template<unsigned long> struct exists; \
    template<typename V> static yes Check (exists<sizeof(&V::NAME)>*); \
    template<typename> static char Check (...); \
    static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes)); \
  }

Instantiate:

HasMember(Foo);

Usage:

HasMember_Foo<B>::value  // without type (but then no overload allowed)
HasMember_Foo<C, int (C::*)(float)>::value  // needs type

Note that, here I am providing two HasMember_Foos, 1 with type and 1 without type. They are generalized for any type (not just specific to int (X::*)(float)). If there is no type mentioned, then the class must have only 1 such method (without overload). Hence, it's always safer to mention the type; As you have done in your question, the specific type is int (X::*)(float). BTW, this also can be included using another macro.

Without such extra macro, in case of class C and class D, you may have to specify the type of the method.

Here is a demo with your code.

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Here it's assumed that whichever class member (function or variable) is chosen, must be public scoped. i.e. If X::foo is private then this solution will not work.

Check if a variable is of function type

Sure underscore's way is more efficient, but the best way to check, when efficiency isn't an issue, is written on underscore's page linked by @Paul Rosania.

Inspired by underscore, the final isFunction function is as follows:

function isFunction(functionToCheck) {
 return functionToCheck && {}.toString.call(functionToCheck) === '[object Function]';
}

Note: This solution doesn't work for async functions, generators or proxied functions. Please see other answers for more up to date solutions.

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