work book

Question

Let S be a stack with operations push, pop, top, empty. What is displayed after following segment of code executes .

                               for(i=1; i<=5; i++)

                               s.push(i);

                               while(!s.empty())

                               printf("%d", s.top());

ans with explanation

Comments

i think we will have a stack overflow because we dont have any condition for decrement of while loop and everytime it will print top of stack.

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After for loop- 

5

4

3

2

1

After while loop-
Printed $\rightarrow 555555...........$
As $s.top()$ just returns top element of the stack without popping it, So while will go in infinite loop.

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@soumya29 : but there is no decrement for while case or you can say there is no popping it only displays top while stack is not empty.

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@arvin. Stopping condition is $!s.empty()$
When stack becomes empty, s.empty() will return true so $!s.empty()$ will return false.

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@soumya29 but how can u say that the top is decrementing as we are not popping

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@arvin. Oh..Yes,, I misread the question. It's s.top(). .
You are right.

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yes, here stack overflow error will occur

rt?

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@soumya29 yes because its not s.pop()

@srestha it will be having overflow

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THANKU

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@arvin the while loop will execute infinite times but should we call the final answer as Stack Overflow? Because the contents of the stack will be fixed to 5,4,3,2,1. The while loop will print infinite times but the stack is not overflowing right?