Why we use top== -1 for implementation of stack using simple array?

I am a newbie in the data structure, I have read the implementation of stack using a simple array, the algorithm for this implementation is below:-

 Stack Operations using Array

A stack can be implemented using array as follows...

Before implementing actual operations, first follow the below steps to create an empty stack.

Step 1: Include all the header files which are used in the program and define a constant 'SIZE' with specific value.

Step 2: Declare all the functions used in stack implementation.

Step 3: Create a one dimensional array with fixed size (int stack[SIZE])

Step 4: Define a integer variable 'top' and initialize with '-1'. (int top = -1)

Step 5: In main method display menu with list of operations and make suitable function calls to perform operation selected by the user on the stack.

push(value) - Inserting value into the stack

In a stack, push() is a function used to insert an element into the stack. In a stack, the new element is always inserted at top position. Push function takes one integer value as parameter and inserts that value into the stack. We can use the following steps to push an element on to the stack...

Step 1: Check whether stack is FULL. (top == SIZE-1)

Step 2: If it is FULL, then display "Stack is FULL!!! Insertion is not possible!!!" and terminate the function.

Step 3: If it is NOT FULL, then increment top value by one (top++) and set stack[top] to value (stack[top] = value).

pop() - Delete a value from the Stack

In a stack, pop() is a function used to delete an element from the stack. In a stack, the element is always deleted from top position. Pop function does not take any value as parameter. We can use the following steps to pop an element from the stack...

Step 1: Check whether stack is EMPTY. (top == -1)

Step 2: If it is EMPTY, then display "Stack is EMPTY!!! Deletion is not possible!!!" and terminate the function.

Step 3: If it is NOT EMPTY, then delete stack[top] and decrement top value by one (top--).

display() - Displays the elements of a Stack

We can use the following steps to display the elements of a stack...

Step 1: Check whether stack is EMPTY. (top == -1)

Step 2: If it is EMPTY, then display "Stack is EMPTY!!!" and terminate the function.

Step 3: If it is NOT EMPTY, then define a variable 'i' and initialize with top. Display stack[ i] value and decrement i value by one (i--).

Step 3: Repeat above step until i value becomes '0'.

Now my questioin is why we use top== -1 in stack instead of  top== 1
Sourodip KunduAsked:
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Julian HansenCommented:
Top is the index into your array to indicate where the top of the stack is in the array.
-1 means the stack is empty.

Arrays are zero indexed so the first element in the stack would be element 0
i.e.
int stack[SIZE];
int top = -1;
int somevalue = 5; // Value to emulate item to be pushed onto stack
// We now add an element. First we check and then increase the stack pointer
if (top < SIZE - 1) {
   top++;
   stack[top] = somevalue;
}
else  {
  // message to say stack is full
}
// top is now equal to 0 - the index of the top of the stack. It was -1 (before the ++ operation) which indicates the stack was empty)

Open in new window

If we pop the stack
top--;
// Top is now -1 indicating that the stack is empty

Open in new window


Checking if top == 1 does not indicate whether the stack is empty - there would have to be two items still in the stack for top to be 1.

It is because arrays are zero indexed that we use the -1 value to determine if it is empty as -1 is not a valid index into the stack array.
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ozoCommented:
If we initialized top=1, then step 3 of push(value) - Inserting value into the stack
 If it is NOT FULL, then increment top value by one (top++) and set stack[top] to value (stack[top] = value).
would push the first value onto stack[2], wasting stack[0] and stack[1], which you'd never get to.

Other than the waste of space, as long as you also changed
Check whether stack is EMPTY. (top == -1)
to
Check whether stack is EMPTY. (top == 1)
then everything would still function, but you'd get to
display "Stack is FULL!!! Insertion is not possible!!!" and terminate the function.
with two fewer push operations than you would have if you had made use of stack[0] and stack[1]
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