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c++c++11lambdarvo

Lambda expressions and RVO

Is the concept of "Return Value Optimization" applied for lambda expression in C++ Compilers? I know that it depends on the compiler and the optimization parameters but is it theoretical possible?

BTW, does anyone know about this issue in VS.NET 2013 or higher?

Solution

  • Yes, it is possible. You can prove it with a little example.

    The following code produced this output, when I compiled with clang and g++ with the -O2 option:

    Ctor

    So, "copy" was not printed. This means that NO copy happened.

    #include <iostream>

class Test
{
public:
    Test() { std::cout << "Ctor\n";}
    Test(const Test& t) 
    {
        std::cout << "copy" << std::endl;
    }
};

int main()
{    
    auto myLambda = []() 
    {
        return Test();
    };

    Test t = myLambda(); 
}

    RVO applies to the return value of a function. A lambda is compiled as a functor. So, it still is a function.

    As for why does it not work in VS, maybe this post can help you.