std::is_bind_expression

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Defined in header <functional>
template< class T >
struct is_bind_expression;
(since C++11)

If T is the type produced by a call to std::bind, this template provides the member constant value equal true. For any other type, value is false.

This template may be specialized for a user-defined type which should be treated by std::bind as if it was the type of a bind subexpression: when a bind-generated function object is invoked, a bound argument of this type will be invoked as a function object and will be given all the unbound arguments passed to the bind-generated object.

Contents

Inherited from std::integral_constant

Member constants

value
[static]
true if T is a function object generated by std::bind, false otherwise
(public static member constant)

Member functions

operator bool
converts the object to bool, returns value
(public member function)

Member types

Type Definition
value_type bool
type std::integral_constant<bool, value>

[edit] Example

#include <iostream>
#include <type_traits>
#include <functional>
 
struct MyBind {
    typedef int result_type;
    int operator()(int a, int b) const { return a + b; }
};
 
namespace std {
    template<>
    struct is_bind_expression<MyBind> : public true_type {};
}
 
int f(int n1, int n2)
{
    return n1+n2;
}
 
int main()
{
    // as if bind(f, bind(MyBind::operator(), _1, _2), 2)
    auto b = std::bind(f, MyBind(), 2); 
 
    std::cout << "Adding 2 to the sum of 10 and 11 gives " << b(10, 11) << '\n';
}

Output:

Adding 2 to the sum of 10 and 11 gives 23

[edit] See also

(C++11)
binds one or more arguments to a function object
(function template)