Subj : chaining cout with a changing object
To   : borland.public.cpp.borlandcpp
From : "Taras Kentrschynskyj" <taras@syd.kth.se>
Date : Wed Sep 10 2003 04:12 pm


One of my students at the Royal Institute of Technology in 
Stockholm, Sweden, draw my attention to the following:
He had written an integer stack class (linked list 
implementation) with a pop operation returning the top item
(data in the first node of the list, deleting this node after
setting the top pointer to it's next), and tried the following:

cout << mystack.pop() << mystack.pop() << mystack.pop();

If he had pushed 1, 2 and 3 on the stack, in that order, the 
output was, unexpectedly:
123
looks like a recursive call.. whereas the following:

cout << mystack.pop();
cout << mystack.pop();
cout << mystack.pop();

of course gave the expected output:
321

I tried to track the first alternative, and found out that 
all three function calls are accomplished before any output
in the chained statement.
The same is true if you use another member function that doesn't 
change the object, e g get(int i), returning item no i in the 
stack. In this case, however, if you chain cout:

cout << mystack.geti(1) << mystack.geti(2) << mystack.geti(3);

the output comes in the chain order, not the reverse order as 
with the pop function.

So, what is going on? How are the three return values saved
in the chained statement? How does the pointer settings affect
the order of the chained output?

.