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?
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.
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;
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.
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.
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.
error: invalid initialization of non-const reference of type ‘int&’ from a temporary of type ‘int’r++returnsint, notint &. Tryr++; return r;instead.r++doesn't "return" anything. It evaluates to a prvalue.