BIO303 Midterm Short Questions By Freshsweb.com
At the beginning, the stack is empty. First, we push the value 2 onto the stack. Like
resulting in the number 2 being placed on the stack. We have a top pointer that points up
element. Then we said push(5). Now look at how the 2 and 5 are stacked. The number 5 is
placed at the top of the number 2 and the top of the pointer moves up one step. Then we are
he pressed the number 7 which is placed at the top and the number 2 and 5 are at the bottom.
Similarly, we press the number 1. The last image in the first line shows the stacked values
of numbers – 1, 7, 5 and 2.
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We remove the elements from the stack. The first picture of the second row shows pop
act. As a result, the number 1 pops up. After that, we press the number 21 again
stack. Numbers 7, 5 and 2 are already in the stack and number 21 is being pushed
top. If we jump now, the number 21 is dropped. Now number 7 is on top. If we did
pop again, number 7 is cracked. Pop again and the number 5 will appear and the number
2 remains in the stack. Here with the help of this diagram we prove that
values are added at the top and removed from the top of the stack.
BIO303 Midterm Short Questions By Freshsweb.com
The last element to go on the stack is the first to come off. That’s why the stack
known as LIFO (Last In First Out) structure. We know that the last element has been squeezed in
the stack is on top, which is removed when we call pop. Let’s look at the next one
scenarios. What happens if we call pop() when there is no element? One possible
the way out is that we have a isEmpty() function that returns true if the stack is empty and false
Otherwise. This is a boolean function that returns true if there is no element in the file
stack. Otherwise, it returns false. The second possibility is that if we call pop
throws an exception on an empty stack. This is an advanced C++ concept.
Exception is also a way to communicate that some unusual condition or
something went wrong. Suppose we have a divide method and try to divide
some number with zero. This method throws a “divide by zero” exception.