Lambda expressions and RVO

Question

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?

RVO is not applied in VS 2013 ? Do you have the flag /O2 which optimize code for speed ? https://msdn.microsoft.com/en-us/library/k1ack8f1.aspx — Pumkko, Oct 19 '15 at 11:00

score 4 · Accepted Answer · edited May 23 '17 at 10:28

4

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.

edited May 23 '17 at 10:28

Community

1
1

answered Oct 19 '15 at 10:57

Pumkko

1,523
15
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sorry I did not get you... how did you exactly make sure by this example that it is applied? – Humam Helfawi Oct 19 '15 at 11:02
The copy constructor is not called only the constructor. So a copy was not made. – Pumkko Oct 19 '15 at 11:03
1

To be meaningful, such code should probably do something with `t` so that the entire program isn't just optimised away to nothing. It _wasn't_ in my case with `g++`, and nor did adding a member `int a{4}` and `cout << t.a` make any difference, so I can confirm your results... but still, when we're measuring optimisations, we should always take steps to make sure _other_ optimisations don't make our measurements meaningless! ;-) – underscore_d Jul 03 '16 at 00:01

score 0 · Answer 2 · answered Oct 19 '15 at 11:02

but is it theoretical possible?

I don't see any reason why not. lamda's are on-the-fly compile time generated structs with () operator overloaded.

meaning that this :

auto f = []{printf("hi");};

will probably be translated to

struct lambda<someID>{  void operator(){printf("hi");}  };
auto f = lambda<someID>();

so, there is no reason why RVO wouldn't work here if the compiler thinks it may optimize things up.

Lambda expressions and RVO

2 Answers2