Trailing return type

In computer programming, a subroutine (a.k.a. function) will often inform calling code about the result of its computation, by returning a value to that calling code. The data type of that value is called the function's return type.

In the C++ programming language, a function must be declared. The C++ function's return type is specified as a part of declaring that function.^[1] A trailing return type, a syntax feature available since C++11, is like a traditional return type, except that it is specified in a different location.^[2][3][4]

Syntax[edit]

An ordinary return type is specified before the function's name. In this example of traditional C++ code, the return type of HasMultipleItems() is bool:

class CClass { public:     bool HasMultipleItems();     std::vector<int> m_veciMember; };  bool CClass::HasMultipleItems() {     return m_veciMember.size() > 1; } 

A trailing return type is specified after the parameter list, following -> symbols:

class CClass { public:     auto HasMultipleItems() -> bool;     std::vector<int> m_veciMember; };  auto CClass::HasMultipleItems() -> bool {     return m_veciMember.size() > 1; } 

Distinction from other language features[edit]

In modern C++, the meaning of the auto keyword will depend on its context:

• When used in a variable's definition (e.g., auto x = 11;), the auto keyword indicates type inference. The data type for that x will be deduced from its initialization. The return type of a function can also be inferred by using auto without specifying a trailing return type (e.g.

  auto CClass::HasMultipleItems() {     return m_veciMember.size() > 1; } 

  )
• On the other hand, there is no type inference in the HasMultipleItems() example on the previous section. That example only uses the auto keyword as a syntactic element, because a trailing return type is being used.

Rationale[edit]

Consider the task of programming a generic version of int Add(const int& lhs, const int& rhs) { return lhs + rhs; }. A proper expression of this function's return type would use the two formal parameter names with decltype: decltype(lhs + rhs). But, where a return type is traditionally specified, those two formal parameters are not yet in scope. Consequently, this code will not compile:

// This will not compile template<typename TL, typename TR> decltype(lhs + rhs) Add(const TL& lhs, const TR& rhs) {     return lhs + rhs; } 

The formal parameters are in scope, where a trailing return type is specified:

template<typename TL, typename TR> auto Add(const TL& lhs, const TR& rhs) -> decltype(lhs + rhs) {     return lhs + rhs; } 

See also[edit]

• Language enhancements of C++ 11

References[edit]