CODEWITHSUNDEEP
c++functionc++11
Yun Huang
Yun Huang

Reputation: 4386

How to get the number of arguments of `std::function`?

Is that possible to get the number of arguments of std::function ? Something like NumOfArgument<...>::value.

For example, NumOfArgument<function<int(int, int)> >::value should be 2.

Upvotes: 16

Views: 6552

Answers (1)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361472

I think std::function itself doesn't provide that feature. But you can implement it yourself as:

template<typename T> 
struct count_arg;

template<typename R, typename ...Args> 
struct count_arg<std::function<R(Args...)>>
{
    static const size_t value = sizeof...(Args);
};

Test code:

typedef std::function<int(int, int)> fun;
std::cout << count_arg<fun>::value << std::endl; //should print 2

See this : Online demo

Likewise, you may put more features into this, as :

template<typename T> 
struct function_traits;     //renamed it!

template<typename R, typename ...Args> 
struct function_traits<std::function<R(Args...)>>
{
    static const size_t nargs = sizeof...(Args);

    typedef R result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
    };
};

Now you can get each argument type, using const index, as:

std::cout << typeid(function_traits<fun>::arg<0>::type).name() << std::endl;
std::cout << typeid(function_traits<fun>::arg<1>::type).name() << std::endl;
std::cout << typeid(function_traits<fun>::arg<2>::type).name() << std::endl;

Working demo

It prints the mangled-name of the types!

Upvotes: 48

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