C++ lambda expression does not compile

Question

I am trying to cin a loop index's value in the loop itself using lambda expression:

#include<iostream>
using namespace std;

int main(){
  for(int a, ([](int & b){cin>>b;})(a); a < 2; ++a);
  return 0;
}

These are the errors when i compile using g++ 4.5 on ubuntu:

forLoopAndCinTest.c++: In function ‘int main()’:
forLoopAndCinTest.c++:5:14: error: expected unqualified-id before ‘[’ token
forLoopAndCinTest.c++:5:14: error: expected ‘)’ before ‘[’ token
forLoopAndCinTest.c++:5:34: error: expected primary-expression before ‘)’ token
forLoopAndCinTest.c++:5:34: error: expected ‘;’ before ‘)’ token
forLoopAndCinTest.c++:5:40: error: name lookup of ‘a’ changed for ISO ‘for’ scoping
forLoopAndCinTest.c++:5:40: note: (if you use ‘-fpermissive’ G++ will accept your code)
forLoopAndCinTest.c++:5:50: error: expected ‘;’ before ‘)’ token

If i use a normal function instead of the lambda, program compiles fine.
Using -fpermissive doesnt help either.
Any ideas?

The current C++ standard doesn't have lambda expressions, so have you specified the option to enable C++0x support? — user180247
– user180247, Commented May 7, 2011 at 21:25
Oh yes Steve. I do use -std=C++0x flag when compiling. Plus, g++ complains if the flag is not used. — badmaash
– badmaash, Commented May 7, 2011 at 21:29
I dont think you capture variables in (). [] is used to capture variables. () is used to pass parameters to the lambda. — badmaash
– badmaash, Commented May 7, 2011 at 21:39

wilhelmtell · Accepted Answer · 2011-05-07 21:37:58Z

5

That's not how the for look works. You are trying to call a lambda where the compiler expects you to declare an int:

for( int a, int2, ...; a < 2; ++a );

Now,

If i use a normal function instead of the lambda, program compiles fine

Yes, but it's probably not doing what you think it does.

void f(int& b)
{
    cin >> b;
}

// ...
for( int a, f(a); a < 2; ++a );

Here, the loop declares two int variables, named a and f. The loop doesn't call f() as you might expect.

Try this instead:

for( int a; cin >> a && a < 2; ++a );

answered May 7, 2011 at 21:37

wilhelmtell

59k20 gold badges98 silver badges131 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

badmaash Over a year ago

I did not understand how the loop 'thinks' that a and f are variables and why f() isnt called. If it is not called, how is the cin statement executed? Thank you for the last idiom.

badmaash Over a year ago

Got it. It doesnt execute the function at all. It just declares a variable with the appropriate type.

badmaash Over a year ago

By the way i got what i wanted: for(int a, x = (cin >> a && 1); a < 10; ++a). Scans the variable inside the for loop although has to use an extra variable and the && operation. But cool. :)

fouronnes · Accepted Answer · 2011-05-07 21:46:26Z

2

The first part of the for is interpreted as a declaration. We get the very same error when replacing your code by the (almost) equivalent :

int main(){
    int a, ([](int & b){cin>>b;})(a); // This produces the same error
    for(; a < 2; ++a);
    return 0;
}

To answer a comment you made, for (int a, foo() ; ... works, but not like you think it does. It is in fact declaring a function (inside the for scope) that returns an int, and has the name foo. As in :

int a, foo();

Which you should read as :

int a;
int foo();

edited May 7, 2011 at 21:46

answered May 7, 2011 at 21:35

fouronnes

4,06826 silver badges46 bronze badges

2 Comments

badmaash Over a year ago

All this is to avoid the cin outside the loop. Yes, this will definitely work. But i still dont understand why i can call any normal function in the declaration part but not lambda.

fouronnes Over a year ago

You can't. When you write for (int a, foo() ; a<2 ; a++) you are NOT calling the foo function. (Try it for yourself).

c-smile · Accepted Answer · 2011-05-07 21:28:10Z

0

After this: for( int a, compiler expects some name (of the variable) - unqualified-id. But in your case it is not so.

answered May 7, 2011 at 21:28

c-smile

27.6k8 gold badges66 silver badges91 bronze badges

1 Comment

c-smile Over a year ago

Yeah, it is just an expression list so your declaration should work either. Interesting.

Collectives™ on Stack Overflow

C++ lambda expression does not compile

3 Answers 3

3 Comments

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related