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Infix to Postfix Conversion Guide

The document describes two algorithms for converting infix notation to postfix notation: 1. A manual algorithm that fully parenthesizes the expression, replaces right parentheses with operators, and removes left parentheses. 2. A stack-based algorithm that reads symbols, outputs operands, pushes operators and parentheses to a stack, and pops operators to the output based on precedence until the entire expression has been processed.

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58 views1 page

Infix to Postfix Conversion Guide

The document describes two algorithms for converting infix notation to postfix notation: 1. A manual algorithm that fully parenthesizes the expression, replaces right parentheses with operators, and removes left parentheses. 2. A stack-based algorithm that reads symbols, outputs operands, pushes operators and parentheses to a stack, and pops operators to the output based on precedence until the entire expression has been processed.

Uploaded by

dangerman
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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CSc 227 — Program Design and Development

(McCann)

Infix → Postfix Conversion Algorithms

1. Manual:

(a) Fully parenthesize the the infix expression (one set of parentheses per operator)
(b) Replace the right parentheses with their corresponding operators
(c) Remove the left parentheses

Example: 4 + 5 * 6
(a) ( 4 + ( 5 * 6 ) )

(b) ( 4 ( 5 6 * +

(c) 4 5 6 * +

2. Stack-based:

while there are more symbols to be read


read the next symbol
case:
operand --> output it
’(’ --> push it on the stack
’)’ --> pop operators from the stack to the output
until a ’(’ is popped; do not output the
parentheses
operator --> pop higher- or equal-precedence operators
from the stack to the output; stop before
popping a lower-precedence operator or
a ’(’. Push the operator on the stack.
end case
end while
pop the remaining operators from the stack to the output

Example: A / (B + C) - D

Input Symbol Stack Content Output


A nil A
/ / A
( /( A
B /( AB
+ /(+ AB
C /(+ ABC
) / ABC+
- - ABC+/
D - ABC+/D
<eof> nil ABC+/D-

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