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Difference between revisions of "cpp/utility/functional/plus"

From cppreference.com
< cpp‎ | utility‎ | functional
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===Member functions===
 
===Member functions===

Revision as of 11:42, 9 June 2018

 
 
Utilities library
General utilities
Relational operators (deprecated in C++20)
 
Function objects
Function invocation
(C++17)(C++23)
Identity function object
(C++20)
Transparent operator wrappers
(C++14)
(C++14)
(C++14)
(C++14)  
(C++14)
(C++14)
(C++14)
(C++14)
(C++14)
(C++14)
(C++14)
(C++14)
(C++14)

Old binders and adaptors
(until C++17*)
(until C++17*)
(until C++17*)
(until C++17*)  
(until C++17*)
(until C++17*)(until C++17*)(until C++17*)(until C++17*)
(until C++20*)
(until C++20*)
(until C++17*)(until C++17*)
(until C++17*)(until C++17*)

(until C++17*)
(until C++17*)(until C++17*)(until C++17*)(until C++17*)
(until C++20*)
(until C++20*)
 
Defined in header <functional>
template< class T >
struct plus;
(until C++14)
template< class T = void >
struct plus;
(since C++14)

Function object for performing addition. Effectively calls operator+ on two instances of type T.

Contents

Specializations

The standard library provides a specialization of std::plus when T is not specified, which leaves the parameter types and return type to be deduced.

function object implementing x + y deducing parameter and return types
(class template specialization) [edit]
(since C++14)

Member types

Type Definition
result_type (deprecated in C++17)(removed in C++20) T
first_argument_type (deprecated in C++17)(removed in C++20) T
second_argument_type (deprecated in C++17)(removed in C++20) T

These member types are obtained via publicly inheriting std::binary_function<T, T, T>.

(until C++11)

Member functions

operator()
returns the sum of two arguments
(public member function) [edit]

std::plus::operator()

T operator()( const T& lhs, const T& rhs ) const;
(until C++14)
constexpr T operator()( const T& lhs, const T& rhs ) const;
(since C++14)

Returns the sum of lhs and rhs.

Parameters

lhs, rhs - values to sum

Return value

The result of lhs + rhs.

Exceptions

(none)

Possible implementation

constexpr T operator()(const T &lhs, const T &rhs) const 
{
    return lhs + rhs;
}