What does this do in C++? []() {}()

Question

I am unit testing for code coverage, making sure that every possible code path is executed by a unit test.

I find that a switch/case element which merely contains a break can be breakpointed, but that the break is never hit, control just jumps to the end of the switch, presumably because of compiler optimization.

A colleague is arguing that that I have not adequately unit test that path.

So, I searched and found an S.O question - which I can not longer find - about C++ code that does nothing. The only answer that didn't also get optimized away of generate compiler or static code analysis errors was []() {}().

This works, insomuch as a breakpoint on it will be hit.

Problem solved, I guess, but what does that actually do?

`[](){}` is a do nothing lambda function. And the extra `()` calls that function. Not sure why that could not be optimised away however. — john, Aug 02 '22 at 06:17
[What does `[](){}` construction mean in c++?](https://stackoverflow.com/q/15389579/995714), [How is `int main(){(([](){})());}` valid C++?](https://stackoverflow.com/q/13603286/995714), [What is a lambda expression in C++11?](https://stackoverflow.com/q/7627098/995714) — phuclv, Aug 02 '22 at 07:02
For your colleague, you could replace it with `<::>()<%%>();` which has the advantage of looking more bug-like. — Eljay, Aug 02 '22 at 12:03
Lol! I like the look of that, but won't pretend that I understand it :-) — Mawg says reinstate Monica, Aug 03 '22 at 07:43

infinitezero · Accepted Answer · 2022-08-02T06:31:08.207

This is an empty lambda that gets executed and does nothing. Usually to set a breakpoint in debugging, at the end of a function or another code block.

[] () {   } ()
^^^ ------------------- no captures
   ^^^ ---------------- no parameters
        ^^^------------ function body
             ^^ ------- call operator (operator()) with no arguments.

Compare this with a Lambda that actually does something:

int x = 5;
auto lambda = [&x](int i){ x=x*i; };
lambda(2);
std::cout << "x is now: " << x << "\n"; // prints x=10

If a function gets optimised out or not is compiler specific.

I composed a small example on godbolt

#include <iostream>

inline void nothing(){}

int main(){
    int x = 5;
    [](){}();
    nothing();
    std::cout << x+3 << "\n";
}

Depending on the optimisation -O0 or -O1, ... and if the nothing() function is marked inline or not, you can see a difference in the assembly.

In gcc12.1 and -O1 the empty lambda is optimised out as well as the call to nothing(). If you remove the inline keyword, even the nothing() declaration disappears as well

Ah, thanks for clarifying. So, that code is, in effect, an infinite zer0? — Mawg says reinstate Monica, Aug 02 '22 at 07:08

What does this do in C++? []() {}()

1 Answers1