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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?
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The lambda function can take many arguments but can return only one expression.
Lambdas can both capture variables and accept input parameters. A parameter list (lambda declarator in the Standard syntax) is optional and in most aspects resembles the parameter list for a function. auto y = [] (int first, int second) { return first + second; };
A return statement is not an expression in a lambda expression. We must enclose statements in braces ({}). However, we do not have to enclose a void method invocation in braces. The return type of a method in which lambda expression used in a return statement must be a functional interface.
Permalink. All the alternatives to passing a lambda by value actually capture a lambda's address, be it by const l-value reference, by non-const l-value reference, by universal reference, or by pointer.
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; } If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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