CODEWITHSUNDEEP
c++static
Itzik984
Itzik984

Reputation: 16764

Function returning static variable C++

I fail to understand why the following program wrong:

 int& getID(){
   static int r = 0;
   return r++;
 }

main:

 int main(){
   int a = getID();
   std::cout << "a=" << a << std::endl;
   return 0;
 }

Why returning a static variable as described creates problems and not returning the wanted value?

Upvotes: 0

Views: 4785

Answers (5)

Alan Stokes
Alan Stokes

Reputation: 18964

Make your function return int not int & and all will be well. You want to return the value of the new id, not a reference to the function's internals.

Upvotes: 1

Vyktor
Vyktor

Reputation: 20997

You should read about prefix and postfix operator and how they are implemented.

Basically, ++i does this (prefix):

i += 1;
return i;

And i++ does (postfix):

ans = i;
i += 1;
return ans;

According to mentioned page, only the prefix operator++ returns reference to upgraded variable. Postfix (i++) returns new variable.

Upvotes: 0

Luchian Grigore
Luchian Grigore

Reputation: 258568

What you're running into is undefined behavior. Anything can happen.

r++ returns a temporary, and it's UB returning temporaries by reference.

On my platform, for example, it doesn't even compile.

Upvotes: 1

Benjamin Lindley
Benjamin Lindley

Reputation: 103693

You are using post-increment(r++ as opposed to ++r). The result of post-increment is a temporary, and you are trying to return a reference to that temporary. You can't do that. If you want to return a reference to r, then you can use pre-increment, or you can just do the increment, then in a separate statement, return r.

Upvotes: 7

Bo Persson
Bo Persson

Reputation: 92231

It doesn't return a reference to r but a reference to r's value before it was incremented. And that is probably lost in action.

Try

r++;
return r;

or

return ++r;

Upvotes: 1

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