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++?

Solution

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)).

Comments

[sehe]

Hey! That's mine. "Don't sweat it. We have int main(){(([](){})());} which is valid C++" (Nov 9th in chat)

[Admin]

it's a c++11 lambda closure

[Ramon Snir]

Read about C++ lambdas, this is what [](){} is (an empty lambda function).

[Mysticial]

6 years of C++, and I've never learned lambdas... I guess that's next on my to-do list.

[sehe]

Also: "@Mysticial int main()??<((??(??)()??<??>)());??>" - much more convincing

[SChepurin]

@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).

[Mysticial]

@sehe My fault for not getting the message. The rest of the room was in favor of posting this trivia question. So I missed your dissenting opinion. I'll give you call on whether or not we should delete this.

[Pete Becker]

@Mysticial - "6 years of C++" -- lambdas were just added in C++11, so nobody has experience with them before a year or so ago.

[tckmn]

The URL here is quite amusing: "how-is-int-main-valid-c"

[Trismegistos]

Is this equation correct {(([ = ](){})());}?

[Avinash R]

just for info, this will compile on gcc (using g++ command). with g++ -std=c++11 file.c

[guest-418]

int main(){ return !<:]()<%;[=:>(){<:&](){;%>();%>();}; } is still valid. I don't even remember where I have found it(somewhere on so).

[Fermat's Little Student]

I used to think JavaScript is the only esoteric programming language widely used in production...

[AnArrayOfFunctions]

It is also valid: int ((main)()){(([](){})());} but I don't brag about it.

[Admin]

N​­​­​­o​­​­​­​­

[Xeo]

@R.MartinhoFernandes: It was still stuck in someone, so I had to go and retrieve it.

[sehe]

I was gonna upvote for correctly mentioning the case where adding () around an expression alters the semantics. But then I remembered that has no relation to the question, really. Nice answer

[Ad N]

The parentheses are also changing the meaning of a program in the case of the most vexing parse disambiguation : B foo(A()) foo is a function (taking a pointer to function as only parameter and returning a B) whereas in B foo((A())) foo is a B object constructed invoking a constructor taking a A object (which instance is an anonymous temporary in this case).

[Xeo]

@AdN: That's not an expression anymore, but a declaration.

[Yakk - Adam Nevraumont]

@R.MartinhoFernandes I think Ramon thought you meant the lambda-dagger, which stabs the return type. (It is usually called an "arrow", but I could see calling -> a "dagger")