Help understanding stacks

[flaca2]

I need help understanding how a stack works in the following code

stack1.push(10)
stack1.push(8)
stack2.push(2)
stack1.push(3)

while !(stack1.isempty())
x=stack1.top;
stack1.pop();
stack1.pop();
stack2.push(x);

My interpretation is I first push all elements into either stack 1 or 2.

So stack 1 would be:
3
8
10

Stack 2 would be
2

Then after the first loop
since Stack1 is not empty x =3
stack1.pop(); will pop 3
stack1.pop(); will pop 8
leaving 10 as x?? and then I push that over to stack 2 leaving stack 1 empty??

I guess the consecutive pop() arguments are making it confusing.

Can anyone help clarify if this correct?

[phoolpreet]

hi guy
i think u got it wrong value of x will not be 10 but 3 becaz we have not change the assignment of x after 2nd stack1.pop()
this will push 3 over 2 in stack2.
in next loop stack1.top will be 1.
so 1st stack1.pop() will pop 10
next stack1.pop() will go in vain and this push 1 in stack2.
loop end here.

[flaca2]

Thank u for the help.
I got a little confused with your statement
"this will push 3 over 2 in stack2.
in next loop stack1.top will be 1."
After it pushes 3 over to stack 2 would stack1.top not be 8 instead of 1?

[lqdeffx]

FILO ( first in last out )
As it was once explained to me one, you are stacking plates so the last thing you place onto your stack, is ontop. so with that said:

stack1 is

10
8
3

and the pop function takes from the top!

[Ganon11]

stack1.push(10)
stack1.push(8)
stack2.push(2)
stack1.push(3)
 
while !(stack1.isempty())
x=stack1.top;
stack1.pop();
stack1.pop();
stack2.push(x);

First of all, this little section of code is an error and will result in an infinite loop. There are no curly brackets around the loop, so only the first statement is executed - x is continually assigned stack1.top, andsince the stack never changes, the loop condition never changes. However, I understand that this is likely a typo, and so I'm assuming you meant to write

while (!(stack1.isEmpty())) {
   x = stack1.top; // should be stack1.top()?
   stack1.pop();
   stack1.pop();
   stack2.push(x);
}

My interpretation is I first push all elements into either stack 1 or 2.

So stack 1 would be:
3
8
10

Stack 2 would be
2

That is correct.

Then after the first loop
since Stack1 is not empty x =3
stack1.pop(); will pop 3
stack1.pop(); will pop 8
leaving 10 as x?? and then I push that over to stack 2 leaving stack 1 empty??

Here, you make two mistakes. In one execution of the loop, x is assigned only once, at the beginning, to stack1.top. That is currently 3. So x is an integer variable holding the value 3. Then you pop two values - 3 and 8 - leaving stack1 to hold 10 and stack2 to hold 2. x still holds 3! It has not been assigned anything else! You then push x (3) onto stack2. So after 1 execution, stack1 holds 10, and stack2 holds 3, 2.

Can you see what will happen in the second execution of the loop?

[phoolpreet]

hi flaca2
let me put in more clear way what i said..
exactly two lines are independent.
we push in stack2 so stack two will be

______
3
-------------
2
_______

stack1.top will be equal to no. of entry in stack 1
stack 1 will look like this


_________
1 <----------------stack1.top = 1(no of entry).
__________