what does <return 0> signifies?

[ankiit]

Hi Mentors,

I wrote a code to implement a small stack about 10 elements, though i read it from a book. I have some issues in understanding the one.

The code is as follows:-

class Stack
{
	int stck[] = new int[10];
	int tos;

	//Constructor

	Stack()
	{
		tos = -1;
	}



//push routine
void push(int item)
	{
	 if(tos >= 9)
		{
			System.out.println("Stack overflow");
		}
	else
		stck[++tos] = item;
	}

//pop rotuine

int pop()
	{
	 if(tos < 0)
		{
			System.out.println("Stack underflow");
			return 0;                                                                   // why return 0??
		}
	else

		return stck[tos--];

	}
}

class Demo
{
	public static void main(String[] args)
	{
		Stack mystack1 = new Stack();
		Stack mystack2= new Stack();

//push items on  to the stack

			for(int i=0; i<10; i++)
				{
					mystack1.push(i);
				}
			for(int j=0; j<10; j++)
				{
					mystack2.push(j);
				}

// pop items from stack

				for(int i=0; i < 10; i++)
					{
						System.out.println(mystack1.pop());
					}

					System.out.println("new line");
					for(int j=0; j < 10; j++)
					{
						System.out.println(mystack2.pop());
					}
			}

}

why we have used return 0, can we use any other thing value with return like return 3.


Thanks in advance
Ankit

[sunde887]

You can use any other value. I don't have all the details of why 0 is used, however; I believe it has to do with being used as a flag indicating failure (some languages -- like c++ use 0 for false, and >0 for true). In the case you have used it you will run into problems. If you have 0 items in a stack and someone attempts to pop the value it should cause an error, since an empty stack can't return anything -- it's empty. Currently it will return 0 when it is empty.

Stack st - new Stack();
int element = st.pop();

This will print that a stack underflow has occurred, but will still be allowed to continue. Which can naturally be problematic. If you have some program with automates over a stack you can cause a problem. The underflow occurs but the automation won't be aware. Instead of printing the error message, I'd suggest you throw an exception.

[ankiit]

Hi sunde887,

I have made some changes to the code, incremented the loop variable from 10 to 11.

Please find is the attached code and the output.

class Stack
{
	int stck[] = new int[10];
	int tos;

	//Constructor

	Stack()
	{
		tos = -1;
	}



//push routine
void push(int item)
	{
	 if(tos >= 9)
		{
			System.out.println("Stack overflow,can't push item onto the stack");
		}
	else
		stck[++tos] = item;
	}

//pop rotuine

int pop()
	{
	 if(tos < 0)
		{
			System.out.println("Stack underflow,can't pop items from stack it's empty");
			return 3;
		}
	else

		return stck[tos--];

	}
}

class Demo
{
	public static void main(String[] args)
	{
		Stack mystack1 = new Stack();
		Stack mystack2= new Stack();

//push items on  to the stack

			for(int i=0; i<11; i++)
				{
					mystack1.push(i);
				}
			/*for(int j=0; j<11; j++)
				{
					mystack2.push(j);
				}  */

// pop items from stack

				//System.out.println("Pop,Stack1");
				for(int i=0; i < 11; i++)
					{
						System.out.println(mystack1.pop());
					}

				/*	//System.out.println("Pop,Stack2");

					for(int j=0; j < 11; j++)
					{
						System.out.println(mystack2.pop());
					}*/
			}

}

And the output is as below:-




Now we can see the return value as 3 in the last line as the output as we have used return 3, or for any matter it shows the value we are using with return.
Still I am unable to understand the purpose of return.

Thanks in advance.
Ankit

[sunde887]

pop is defied to return an int, so it must return one. Try leaving out the return statement and see if it changes.

[ankiit]

Hi sunde887,

The code while compiling throws an error, that the return statement is missing.

Thanks
Ankit

[ankiit]

Hi subnde887,

can we initialize tos to some other value like 10?(10 is 1 above the maximum value set for the stack)?