"Thomas Ibbotson" <> wrote in message
news:dgn2en$g27$
> I've been going through a tutorial on C++ and I've come across what I
> regard as some strange behaviour. As part of the tutorial on classes a
> simple stack is implemented with functions push() and pop().
>
> I test this stack using:
> stack a_stack; // Create an instance of stack
>
> // Push some data onto the stack
> a_stack.push(1);
> a_stack.push(2);
> a_stack.push(3);
>
> // Now pop it off again and see what we get
> cout << a_stack.pop() << " " << a_stack.pop() << " " <<
> a_stack.pop() << endl;
>
> // Refill the stack
> a_stack.push(1);
> a_stack.push(2);
> a_stack.push(3);
>
> // Try it a different way
> cout << a_stack.pop() << endl;
> cout << a_stack.pop() << endl;
> cout << a_stack.pop() << endl;
>
> The resulting output is:
> 1 2 3
> 3
> 2
> 1
>
> Where I would have expected:
> 3 2 1
> 3
> 2
> 1
>
> Can someone tell me what is going on?
>
> Thanks,
> Tom
To better understand the output, run the following code:
int main()
{
stack a_stack; // Create an instance of stack
int a, b, c; // used for recording results of pop operations
// Push some data onto the stack
a_stack.push(1);
a_stack.push(2);
a_stack.push(3);
// Now pop it off and see what we get
cout << (a=a_stack.pop()) << " " << (b=a_stack.pop()) << " "
<< (c=a_stack.pop()) << endl;
cout << "a is " << a << endl;
cout << "b is " << b << endl;
cout << "c is " << c << endl;
// Refill
a_stack.push(1);
a_stack.push(2);
a_stack.push(3);
// Pop it off a different way
cout << endl;
cout << (a=a_stack.pop()) << endl;
cout << (b=a_stack.pop()) << endl;
cout << (c=a_stack.pop()) << endl;
cout << "a is " << a << endl;
cout << "b is " << b << endl;
cout << "c is " << c << endl;
return 0;
}
The output is:
1 2 3
a is 1
b is 2
c is 3
3
2
1
a is 3
b is 2
c is 1
For the first (anomalous) result, what the output tells you is that the
numbers are being output in left-to-right order, as you would expect.
However, prior to doing the outputting, the program is making the pop()
calls in the reverse order to what you expect. Specifically, it is doing
them in the following order:
c=a_stack.pop()
b=a_stack.pop()
a=a_stack.pop()
The reason for this is that an implementation is free to make these
evaluations in any order it wants since all pop() functions are part of the
same expression.
For the second version, the semi-colon at the end of each line creates a
"sequence point" which forces the evaluation of the functions up to that
point. This forces evaluation in the expected order:
a=a_stack.pop()
b=a_stack.pop()
c=a_stack.pop()
--
John Carson
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