what is wrong here? associativity? evaluation order? how to change order?

Question

Hi...

The associativity of stream insertion operator is rtl, forgetting this fact sometimes cause to runtime or logical errors. for example:

1st-

int F()
{
   static int internal_counter c=0;
   return ++c;
}

in the main function:

//....here is main()
cout<<”1st=”<<F()<<”,2nd=”<<F()<<”,3rd=”<<F();

and the output is:

1st=3,2nd=2,3rd=1

that is different from what we expect at first look.

2nd- suppose that we have an implementation of stack data structure like this:

    //
    //... a Stack<DataType> class …… 
    //

    Stack<int> st(10);
    for(int i=1;i<11;i++)
       st.push(i);

cout<<st.pop()<<endl<<st.pop()<<endl<<st.pop()<<endl<<st.pop()<<endl;

expected output is something like:

10
9
8
7

but we have:

7
8
9
10

There is no internal bug of << implementation but it can be so confusing... and finally[:-)] my question: is there any way to change associativity of an operator by overloading it?

do you think this could be not reverse? i mean is it possible to change order by modifying or changing an open source STL?

Answer 1

No there isn't. But I think you may be mixing up associativity with evaluation order. The only operators that specify an evalualtion order are &&, || and , (comma). When you say:

cout<<st.pop()<<endl<<st.pop()<<endl<<st.pop()<<endl<<st.pop()<<endl;

the compiler can evaluate sub-expressions such as st.pop() in any order it likes, which is what causes the unexpected output.

Answer 2

The only things that are right-associative are the assignment operators. See §5.4 to 5.18 of the standard. The << operators are evaluated left-to-right or the messages would be backward in grammar, not in content. The content is due to side effects, which are unordered in C++ except (as Neil mentions) for "short-circuit" && and ||, and comma.

Answer 3

To see how this is an order of evaluation issue and not an associativity issue, modify your code to this:

int a = st.pop();
int b = st.pop();
int c = st.pop();
cout << a << endl << b << endl << c << endl;