How is “int main(){(([](){})());}” valid C++?

Question

I recently came across the following esoteric piece of code.

int main(){(([](){})());}

Reformat it as follows to make it more readable:

int main(){
    (([](){})());   //  Um... what?!?!
}

But I can't get my head around how (([](){})()) is valid code.

It doesn't look like function pointer syntax.
It can't be some operator overloading trick. The code compiles as is.

Google didn't help much with this all-symbol search. But it compiles in Visual Studio 2010 and outputs nothing. There were no errors, and no warnings. So it looks like valid code.

I've never seen any valid code that is so bizarre outside of Javascript and C function pointers.

Can someone explain how this is valid C++?

Hey! That's mine. "Don't sweat it. We have int main(){(([](){})());} which is valid C++" (Nov 9th in chat) — sehe
@Mysticial - This code mystifies you because it is useless. If this lambda would do something, you would recognize it having syntax similar to function pointers (with which it is related closely). — SChepurin
@Mysticial - "6 years of C++" -- lambdas were just added in C++11, so nobody has experience with them before a year or so ago. — Pete Becker

Xeo Xeo · Accepted Answer · 2012-11-28T10:51:48

The code essentially calls an empty lambda.

Let's start from the beginning: [](){} is an empty lambda expression.

Then, in C and C++, you can wrap expressions in parens and they behave exactly the same^† as if written without them, so that's what the first pair of parens around the lambda does. We're now at ([](){}).

Then, () after the first wrapping parens calls the (empty) lambda. We're now at ([](){})()

The whole expression is wrapped in parens again and we get (([](){})()).

At last, ; ends the statement. We arrive at (([](){})());.

† There are some corner cases at least in C++, like with T a_var; there's a difference between decltype(a_var) and decltype((a_var)).

How is “int main(){(([](){})());}” valid C++?

1 Answers