I have a question when it comes to - / operators in postfix vs infix.
From the assignment
The input string 5 4 + 3 10 * + is equivalent to the infix expression (5 + 4) + (3 * 10) The answer is 39.
I follow that. Then I get confused by this statement.
We also have to worry about the non-commuting operators – and / . We will evaluate the postfix string 4 5 – as 4 – 5 and, likewise, will evaluate 4 5 / as 4 / 5 .
When I do that however...I get different results with infix vs postfix.
Modifying the first example to include subtraction.
infix
(5 - 4) + (3 * 10) = 31
postfix
5 4 - 3 10 * +
29....right?
SO I'm confused. The results of infix and postfix are supposed to be the same right? Is this a typo in the actual assignment or am I doing something wrong?
When you're evaluating postfix you use a stack: you push operands and when you get to an operator you pop the required operands and push the evaluated result.
In the case of commutative operators like + it doesn't matter which order the operands are in. So for example:
5 4 +
can be evaluated as
PUSH 5
PUSH 4
PUSH (POP + POP)
where the first POP will yield 4 and the second POP 5. So you have really evaluated 4+5.
But in the case of non-commutative operators, this won't work. You have to evaluate 5 / 4, not 4 / 5. So you need either to use temporary variables:
PUSH 5
PUSH 4
let d = POP; // divisor = 4
let q = POP; // quotient = 5
PUSH q/d; // push the dividend
or else introduce a SWAP operation, which swaps the top two items on the stack:
PUSH 5
PUSH 4
SWAP
PUSH (POP / POP)
or else compile the postfix so as to push in the opposite order:
PUSH 4
PUSH 5
PUSH (POP/POP)