C++

Trailing return type

Syntax#

  • function_name ( [function_args] ) [function_attributes] [function_qualifiers] -> trailing-return-type [requires_clause]

Remarks#

The above syntax shows a full function declaration using a trailing type, where square brackets indicate an optional part of the function declaration (like the argument list if a no-arg function).

Additionally, the syntax of the trailing return type prohibits defining a class, union, or enum type inside a trailing return type (note that this is not allowed in a leading return type either). Other than that, types can be spelled the same way after the -> as they would be elsewhere.

Avoid qualifying a nested type name

class ClassWithAReallyLongName {
  public:
    class Iterator { /* ... */ };
    Iterator end();
};

Defining the member end with a trailing return type:

auto ClassWithAReallyLongName::end() -> Iterator { return Iterator(); }

Defining the member end without a trailing return type:

ClassWithAReallyLongName::Iterator ClassWithAReallyLongName::end() { return Iterator(); }

The trailing return type is looked up in the scope of the class, while a leading return type is looked up in the enclosing namespace scope and can therefore require “redundant” qualification.

Lambda expressions

A lambda can only have a trailing return type; the leading return type syntax is not applicable to lambdas. Note that in many cases it is not necessary to specify a return type for a lambda at all.

struct Base {};
struct Derived1 : Base {};
struct Derived2 : Base {};
auto lambda = [](bool b) -> Base* { if (b) return new Derived1; else return new Derived2; };
// ill-formed: auto lambda = Base* [](bool b) { ... };

This modified text is an extract of the original Stack Overflow Documentation created by the contributors and released under CC BY-SA 3.0 This website is not affiliated with Stack Overflow