The algorithm is: 


For each item in the postfix expression from the left: 

if the item is a number push it on the stack 

if the item is an operator (+,-,*, or /) then 

pop two numbers off the stack 

make a calculation: the second number popped-operator-first number 

push the result on the stack 

When the loop is done the answer is the only item remaining on the stack. 

Parsing an Infix Expression with No Parentheses (like 2 * 3 - 48/4 -4 * 5)
Two stacks are required, one for numbers and the other for operators. The algorithm is: 


For each item in the infix expression (no parens) from the right to the left 

If the item is a number then push it on the number stack. 

If the item is an operator (+,-,*, or /) and: the operator stack is empty or the operator on the top of the stack is higher in priority (* and / are higher in priority than + or -), then 

Pop an operator from the operator stack. 
Pop two numbers off the number stack. 
Calculate using second number-operator-first number. 
Push the result on the number stack. 
Push the item on the operator stack. 

Else push the item on the operator stack. 

After the loop, while the operator stack is not empty 

Pop an operator from the operator stack. 
Pop two numbers off the number stack. 
Calculate using second number-operator-first number. 
Push the result on the number stack. 

The answer is the last item in the number stack. 

Parsing an Infix Expression with Parentheses (like ((2*4-6/3)*(3*5+8/4))-(2+3))
One stack is required. The algorithm is: 


Have a string available to store items 

For each item in the infix (with parens) expression 

If the item is a number then add it to the string with a space 

if the item is an operator then 

While the stack is not empty and an operator is at the top and the operator at the top is higher priority that the item then 
= Pop the operator on the top of the stack 

= Add the popped operator to the string with a space 


Push the item on the stack 

else if the item is a left paren 

Push the item on the stack 

else if the item is a right paren 

Pop a thing off of the stack. 

while that thing is not a left paren 
= Add the thing to the string with a space 

= Pop a thing off of the stack 


While the stack is not empty 

Pop a thing off the stack. 

Add it to the string with a space. 

Remove the last character (a space) from the string.