std::is_bind_expression
From cppreference.com
< cpp | utility | functional
| Defined in header <functional>
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| 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 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; } } my_2; namespace std { template<> struct is_bind_expression<MyBind> : public true_type {}; } int f(int n1, int n2) { return n1+n2; } int main() { auto b = std::bind(f, MyBind(), 2); // as if bind(f, bind(MyBind::operator(), _1, _2), 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) |
