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;
), theauto
keyword indicates type inference. The data type for thatx
will be deduced from its initialization. The return type of a function can also be inferred by usingauto
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 theauto
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]References
[edit]- ^ Stroustrup, Bjarne (2013). The C++ Programming Language (Fourth ed.). Addison-Wesley. ISBN 978-0-321-56384-2.
- ^ "Function declaration". cppreference.com. C++ Reference. Retrieved 1 March 2021.
- ^ "C++0x Suffix Return Types". cplusplus.com. The C++ Resources Network. Retrieved 1 March 2021.
- ^ "Functions (C++)". Microsoft C++, C, and Assembler. Microsoft Corporation. Retrieved 1 March 2021.