This is really weird..

I just wrote a a stack implementation using linked list..
when i pop the values it is as expected

Code:

	stack.push(7);
	stack.push(8) ;
	stack.push(9) ;
	cout << stack.pop()  << endl << stack.pop() << endl ;

when I pop the values..it prints out as 7 8 9! it only prints correctly if I put the print statements on different lines..literally. It doesn't make sense to me..
i debugged and the first pop definitely returns a 9..so how does it end up being at the end?
here is the code.. .cpp file

Code:

#include "StackLinkedList.h"
Stack::Stack()
{
	size=0;
	stackPtr = NULL;
}
Stack::~Stack()
{
	delete stackPtr ;
}

void Stack::push(int x)
{
	++size ;

	Node *temp ;
	temp= stackPtr;
	stackPtr = new Node(x) ;
	stackPtr->next =temp;
}
bool Stack::empty()
{

	return (stackPtr==NULL) ;

}
int Stack::pop()
{
	
		int n;
		Node *t = stackPtr ;
		stackPtr = stackPtr->next ;
		n = t->data;

		delete t ;
		return n ;


}

It's because your program has undefined behaviour.

You're modifying the same object (your stack) more than once between sequence points.

It's like the old
cout << a++ << a++ << endl;
and wondering why the result is one of many possible things you didn't expect.

so there is a bug in the code..aargh, how will I find it..I can't spot anything!

One thing that i noticed is;
You are increasing the size of linked list everytime you push but you don't decrease it when you pop out...

You do
a = pop();
b = pop();
cout << a << b;
so that you introduce some sequence points into the code to make it explicit what is happening.

Originally Posted by Salem

You do
a = pop();
b = pop();
cout << a << b;
so that you introduce some sequence points into the code to make it explicit what is happening.

yep, I did that and that works as well...
but it should be able to chain the function calls return value..without the behavior being undefined..
is it the problem with my code..or a cout issue which i doubt..

Originally Posted by Eman

yep, I did that and that works as well...
but it should be able to chain the function calls return value..without the behavior being undefined..
is it the problem with my code..or a cout issue which i doubt..

You can use two functions in the same "sequence points", as long as they don't have side-effects that influence each other. Look at this:
Sequence point - Wikipedia, the free encyclopedia

Definitely, it's an issue with your written code...

It's undefined - there is nothing more to say.

It's not specific to cout, you could do this, and still be in the same trouble.
answer = pop() / pop();
There is no way for you to know which number is going to get divided by which number.

It's just the way the language works.

brilliant..i guess i have to read the Sequence point stuff
so Mr 777 how do you suggest I fix it..because this was the algorithm I was given..

First, try to understand the concept of stack. (If you don't know)
Second, try to understand your problem specification, i mean what exactly you want to do.
Thrid, take a copy and pen, start making design of your class, objects and different operations. I mean check all cases that what your program should do in order to solve given problem in all scenarios, you can come up with.
Finally, write code, and test it with all possible scenarios you've made while designing and then take some other scenarios (try to crash your written program in every possible manner).
If still problem exists, throw it here and we all will try our best to do what we could.

So, i'll recommend you to write your own code. Belive me, once you understood it well, it'll not take more than 10 minutes to you to develop a working stack class but good designing and understanding is the must thing....

Good luck.

em...thanks for the advice, but it doesn't really help.
I wrote this code.. And I don't write code unless I don't understand the algorithm..I have read about the stack..I have watched videos on stack..and I have accidentally found a bug in the freaking code...although that doesn't tell me where it is or how to remove it

ah screw it.. i will use Salem's idea

Simply don't call pop() more than once in a single expression (where the order matters).

Code:

cout << stack.pop() << '\n';
cout << stack.pop() << '\n';

Normally you use that function in a loop anyway.

Code:

while (!stack.empty()) 
    cout << stack.pop() << '\n';