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

Here is a solution simpler than Johannes Schaub - litb's one. It requires C++11.

#include <type_traits>

template <typename T, typename = int>
struct HasX : std::false_type { };

template <typename T>
struct HasX <T, decltype((void) T::x, 0)> : std::true_type { };

Update: A quick example and the explanation on how this works.

For these types:

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

we have HasX<A>::value == true and HasX<B>::value == false. Let's see why.

First recall that std::false_type and std::true_type have a static constexpr bool member named value which is set to false and true, respectively. Hence, the two templates HasX above inherit this member. (The first template from std::false_type and the second one from std::true_type.)

Let's start simple and then proceed step by step until we get to the code above.

1) Starting point:

template <typename T, typename U>
struct HasX : std::false_type { };

In this case, there's no surprise: HasX derives from std::false_type and hence HasX<bool, double>::value == false and HasX<bool, int>::value == false.

2) Defaulting U:

// Primary template
template <typename T, typename U = int>
struct HasX : std::false_type { };

Given that U defaults to int, Has<bool> actually means HasX<bool, int> and thus, HasX<bool>::value == HasX<bool, int>::value == false.

3) Adding a specialization:

// Primary template
template <typename T, typename U = int>
struct HasX : std::false_type { };

// Specialization for U = int
template <typename T>
struct HasX<T, int> : std::true_type { };

In general, thanks to the primary template, HasX<T, U> derives from std::false_type. However, there exists a specialization for U = int which derives from std::true_type. Therefore, HasX<bool, double>::value == false but HasX<bool, int>::value == true.

Thanks to the default for U, HasX<bool>::value == HasX<bool, int>::value == true.

4) decltype and a fancy way to say int:

A little digression here but, please, bear with me.

Basically (this is not entirely correct), decltype(expression) yields the type of expression. For instance, 0 has type int thus, decltype(0) means int. Analogously, 1.2 has type double and thus, decltype(1.2) means double.

Consider a function with this declaration:

char func(foo, int);

where foo is some class type. If f is an object of type foo, then decltype(func(f, 0)) means char (the type returned by func(f, 0)).

Now, the expression (1.2, 0) uses the (built-in) comma operator which evaluates the two sub-expressions in order (that is, first 1.2 and then 0), discards the first value and results in the second one. Hence,

int x = (1.2, 0);

is equivalent to

int x = 0;

Putting this together with decltype gives that decltype(1.2, 0) means int. There's nothing really special about 1.2 or double here. For instance, true has type bool and decltype(true, 0) means int as well.

What about a class type? For instace, what does decltype(f, 0) mean? It's natural to expect that this still means int but it might not be the case. Indeed, there might be an overload for the comma operator similar to the function func above that takes a foo and an int and returns a char. In this case, decltype(foo, 0) is char.

How can we avoid the use of a overload for the comma operator? Well, there's no way to overload the comma operator for a void operand and we can cast anything to void. Therefore, decltype((void) f, 0) means int. Indeed, (void) f casts f from foo to void which basically does nothing but saying that the expression must be considered as having type void. Then the built-in operator comma is used and ((void) f, 0) results in 0 which has type int. Hence, decltype((void) f, 0) means int.

Is this cast really necessary? Well, if there's no overload for the comma operator taking foo and int then this isn't necessary. We can always inspect the source code to see if there's such operator or not. However, if this appear in a template and f has type V which is a template parameter, then it's no longer clear (or even impossible to know) whether such overload for the comma operator exists or not. To be generic we cast anyway.

Bottom line: decltype((void) f, 0) is a fancy way to say int.

5) SFINAE:

This is a whole science ;-) OK I'm exagerating but it's not very simple either. So I'll keep the explanation to the bare minimum.

SFINAE stands for Substitution Failure is Not An Error. It means that when a template parameter is substituted by a type, an illegal C++ code might appear but, in some circunstances, instead of aborting compilation the compiler simply ignores the offending code as if it wasn't there. Let's see how it applies to our case:

// Primary template
template <typename T, typename U = int>
struct HasX : std::false_type { };

// Specialization for U = int
template <typename T>
struct HasX <T, decltype((void) T::x, 0)> : std::true_type { };

Here, again, decltype((void) T::x, 0) is a fancy way to say int but with the benefit of SFINAE.

When T is substituted with a type, an invalid construct might appear. For instance, bool::x is not valid C++, so substituting T with bool in T::x yields an invalid construct. Under the SFINAE principle, the compiler doesn't reject the code, it simply ignores (parts of) it. More precisely, as we have seenHasX<bool> means actually HasX<bool, int>. The specialization for U = int should be selected but, while instantiating it, the compiler finds bool::x and ignores the template specialization altogether as if it didn't exist.

At this point, the code is essencially the same as in case (2) above where just the primary template exists. Hence, HasX<bool, int>::value == false.

The same argument used for bool holds for B since B::x is an invalid construct (B has no member x). However, A::x is OK and the compiler sees no issue in instantiating the specialization for U = int (or, more precisely, for U = decltype((void) A::x, 0)). Hence, HasX<A>::value == true.

6) Unnaming U:

Well, looking at the code in (5) again, we see that the name U is not used anywhere but in its declaration (typename U). We can then unname the second template argument and we obtain the code shown at the top of this post.

We can use a C++20 requires expression to solve this problem. h/t to @lefticus who recently posted this method in C++ Weekly - Ep 242 - Design By Introspection in C++20 (concepts + if constexpr:

#include <iostream>

struct P1 {int x;};
struct P2 {float X;};

bool has_x(const auto &obj) {
    if constexpr (requires {obj.x;}) {
      return true;
    } else
      return false;
}

int main()
{
    P1 p1 = {1};
    P2 p2 = {1};

    std::cout << std::boolalpha << has_x(p1) << "\n"; 
    std::cout << has_x(p2) << "\n"; 

    return 0;
}

You can see it live here.

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.

I got redirected here from a question which has been closed as a duplicate of this one. I know it's an old thread, but I just wanted to suggest an alternative (simpler?) implementation that works with C++11. Supposing we want to check whether a certain class has a member variable called id:

#include <type_traits>

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

template<typename T>
struct has_id<T, decltype(std::declval<T>().id, void())> : std::true_type { };

That's it. And here is how it would be used (live example):

#include <iostream>

using namespace std;

struct X { int id; };
struct Y { int foo; };

int main()
{
    cout << boolalpha;
    cout << has_id<X>::value << endl;
    cout << has_id<Y>::value << endl;
}

Things can be made even simpler with a couple of macros:

#define DEFINE_MEMBER_CHECKER(member) \
    template<typename T, typename V = bool> \
    struct has_ ## member : false_type { }; \
    template<typename T> \
    struct has_ ## member<T, \
        typename enable_if< \
            !is_same<decltype(declval<T>().member), void>::value, \
            bool \
            >::type \
        > : true_type { };

#define HAS_MEMBER(C, member) \
    has_ ## member<C>::value

Which could be used this way:

using namespace std;

struct X { int id; };
struct Y { int foo; };

DEFINE_MEMBER_CHECKER(foo)

int main()
{
    cout << boolalpha;
    cout << HAS_MEMBER(X, foo) << endl;
    cout << HAS_MEMBER(Y, foo) << endl;
}