Can I force a C++11 lambda to return by reference?

Question

This does not compile since the lambda expression returns by value:

#include <iostream>  class Item { public:     int& f(){return data_;} private:     int data_ = 0; };  int main() {     Item item;     auto lambda = [](Item& item){return item.f();};     lambda(item) = 42;  // lambda(item) is a rvalue => compile time error      std::cout << item.f() << std::endl;     return 0; } 

Is there a way around this? Can I force a lambda to return by reference?

Answer 1

You should specify the lambda return type to be int&. If you leave the return type off [and the lambda is of form return expression; it will automatically deduce the return type.

#include <iostream>  class Item { public:     int& f(){return data_;} private:     int data_ = 0; };  int main() {     Item item;     auto lambda = [](Item& item) ->int& {return item.f();}; // Specify lambda return type     lambda(item) = 42;     std::cout << item.f() << std::endl;     return 0; }