Stacks

	First in, last out processing
	
	You can only see the next available item on the stack, the newest one in the stack
	
	Examples:
	
		your inbox at work
		a pile of dishes in the sink
	
Push

	Add a new item to the top of the stack

	What are the possible errors involved?

Pop

	Remove the top item from the stack
	
	What are the possible errors involved?

Peek

	Look at the top item on the stack.
	
	What are the possible errors involved?

As a Linked List:

	Keep a pointer to the last node of the list
	Push: Add a new node to the end of the list
	Pop : move the ptrLast back one and delete the last node

As an Array:

	Keep the index value of the most recently added item
	Push: increment the current index value
	Pop : decrement the current index value

WriteBackwards

	string wordFwd = char[10];
	string wordBwd = char[10];
	char   chr     = 0;
	
	
	for ( int i=0; i<10; i++ ) {
		push(wordFwd[i]);
	}

	for ( int i=0; i<10; i++ ) {
		chr = pop();
		cout << chr;
	}

Infix Notation

	Operators are placed between operands - the way we normally see them
	
		2 * 5 + 7
		
		postfix string:  2 5 * 7 +
		opertator stack: 


Postfix Notation

	Operators are placed after the operands
	
		257*+
	
Infix to Postfix

	The pseudo-code for converting is:
	
		while (there are more items) {
			read next item into input
			if ( input is identifier or constant )
				append the popped operand to postfix string
			else {
				switch ( input ) {
					case '(':
						push input onto operator stack
					case ')':
						while ( top of operator stack != '(' ) {
							pop the operator stack
							append the popped operand to postfix string
						}
						pop the operator stack
					default:
						while ( input has less or equal priority to top of operator stack ) {
							pop the operator stack
							append the popped operand to postfix string
						}
						push input onto operator stack
				}
			}
		}

Evaluating Postfix Expressions

	This requires two stacks - operand and operator
	
	The pseudo-code is:
	
		while (there are more items) {
			read next item into input
			if ( input is identifier or constant )
				push onto operand stack
			else if ( there are more than two operands on operand stack ) {
				pop top two operands
				apply operator
				push the result onto operand stack
			} else 
				push onto operator stack
		}



		12 25 7 * +
		
		25*7+


		operand stack : 17
		
		operator stack: