Reputation: 13343
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?
Upvotes: 34
Views: 8592
Reputation: 9852
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;
}
Upvotes: 59