How to check if the template parameter of the function has a certain type?

Say I have a function with template type T and two other classes A and B.

template <typename T>
void func(const T & t)
{
    ...........
    //check if T == A do something
    ....开发者_如何学C.......
    //check if T == B do some other thing
}

How can I do these two checks (without using Boost library)?

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If you literally just want a boolean to test whether T == A, then you can use is_same, available in C++11 as std::is_same, or prior to that in TR1 as std::tr1::is_same:

const bool T_is_A = std::is_same<T, A>::value;

You can trivially write this little class yourself:

template <typename, typename> struct is_same { static const bool value = false;};
template <typename T> struct is_same<T,T> { static const bool value = true;};

Often though you may find it more convenient to pack your branching code into separate classes or functions which you specialize for A and B, as that will give you a compile-time conditional. By contrast, checking if (T_is_A) can only be done at runtime.

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Create function templates with specializations, that will do the thing you want.

template <class T>
void doSomething() {}

template <>
void doSomething<A>() { /* actual code */ }

template <class T>
void doSomeOtherThing() {}

template <>
void doSomeOtherThing<B>() { /* actual code */ }

template <typename T>
void func(const T & t)
{
    ...........
    //check if T == A do something
    doSomething<T>();
    ...........
    //check if T == B do some other thing
    doSomeOtherThing<T>();
}

---

If you want to have a special implementation of func for some parameter type, simply create an overload specific for that type:

template <typename T>
void func(const T & t) {
   // generic code
}

void func(const A & t) {
   // code for A
}