Solved-Assignment #6 -Solution

Description

(Infix-to-Postfix Converter) Compilers to help in the process of evaluating expressions and generating machine-language code use stacks. In this exercise, we investigate how compilers evaluate arithmetic expressions consisting only of constants, operators and parentheses. Humans generally write expressions like 3 + 4 and 7 / 9 in which the operator (+ or / here) is written between its operands – this is called infix notation. Computers “prefer” postfix notation, in which the operator is written to the right of its two operands. The preceding infix expressions would appear in postfix notation as 3 4 + and 7 9 /, respectively.

To evaluate a complex infix expression, a compiler would first convert the expression to postfix notation and evaluate the postfix version. Each of these algorithms requires only a single left-to-right pass of the expression. Each algorithm uses a stack object in support of its operation, but each uses the stack for a different purpose. In this exercise, you’ll write a Java version of the infix-to-postfix conversion algorithm. Write class InfixToPostfixConverter to convert an ordinary infix arithmetic expression (assume a valid expression is entered) with single-digit integers such as

(6 + 2) * 5 – 9 / 4

to a postfix expression. The postfix version (no parentheses are needed) of the this infix expression is

6 2 + 5 * 9 4 / –

The program should read the expression into StringBuffer infix and use one of the stack collection classes (i.e., Stack<T>) to help create the postfix expression in StringBuffer postfix. The algorithm for creating a postfix expression is as follows:

1. Push a left parenthesis ‘(‘ onto the stack.

2. Append a right parenthesis ‘)’ to the end of infix.

3. While the stack is not empty, read infix from left to right and do the following:

o If the current character in infix is a digit, append it to postfix.

o If the current character in infix is a left parenthesis, push it onto the stack.

o If the current character in infix is an operator:

 Pop operators (if there are any) at the top of the stack while they have equal or higher precedence than the current operator, and append the popped operators to postfix.

 Push the current character in infix onto the stack.

o If the current character in infix is a right parenthesis:

 Pop operators from the top of the stack and append them to postfix until a left parenthesis is at the top of the stack.

 Pop (and discard) the left parenthesis from the stack.

The following arithmetic operations are allowed in an expression:

+ addition

– subtraction

* multiplication

/ division

^ exponentiation

% remainder

The stack should be maintained with stack nodes that each contain an instance variable and a reference to the next stack node. Some methods you may want to provide are as follows:

1. Method convertToPostfix, which converts the infix expression to postfix notation.

2. Method isOperator, which determines whether c is an operator.

3. Method precedence, which determines whether the precedence of operator1 (from the infix expression) is less than, equal to or greater than that of operator2 (from the stack). The method returns true if operator1 has lower precedence than operator2. Otherwise, false is returned.

4. Method peek, which returns the top value of the stack without popping the stack.

Requirements

Realize your implementation that satisfies the following minimal requirements:

• 2 pt: properly exhibits right logic, i.e., readable and compilable coding

• 2 pt: properly reads infix notation

• 2 pt: properly tokenizes expression

• 4 pt: properly produces postfix notation