initialization by lambda expression in c++

Question

i write follow code：

static int count = []()->int
                   {
                       int count = 0;

                       for(int i = 0; i < categories.size(); ++i)
                       {
                           if(!categories[i].isCategory())
                           {
                               count++;
                           }
                       }

                       return count;
                   };

and got error:error: cannot convert '__lambda0' to 'int' in initialization.

does the meaning of my code fragment is assign the __lambda0 to static int count instead of return the inner count?

Yes, you're assigning the lambda (function), not its evaluation. You need to actually invoke the lambda. — heinrichj
– heinrichj, Commented Feb 28, 2014 at 7:12

Mark Garcia · Accepted Answer · 2014-02-28 07:11:35Z

8

You aren't calling it! Make sure you do so:

static int count = []()->int
                   {
                       int count = 0;

                       for(int i = 0; i < categories.size(); ++i)
                       {
                           if(!categories[i].isCategory())
                           {
                               count++;
                           }
                       }

                       return count;
                   }();
                 // ^^ THIS THIS THIS THIS

BUT, IMHO, you're better off in this without using a lambda. And in the case where you'd use it in other parts of your code, then have it in a stand-alone (not lambda) function.

answered Feb 28, 2014 at 7:11

Mark Garcia

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1 Comment

Potatoswatter Over a year ago

I think this is a valid use-case. It might be a static local, and the lambda could depend on other locals. (But this example doesn't show anything get captured.)

Potatoswatter · Accepted Answer · 2014-02-28 07:11:42Z

2

does the meaning of my code fragment is assign the __lambda0 to static int count instead of return the inner count?

Exactly. To call the lambda, just add () at the end.

                   …
               } ();

answered Feb 28, 2014 at 7:11

Potatoswatter

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