Arithmetic notation conversion infix to1
- 1. Arithmetic Expressions • Infix form – operand operator operand • 2+3 or • a+b – Need precedence rules – May use parentheses • 4*(3+5) or • a*(b+c)
- 2. Arithmetic Expressions • Postfix form – Operator appears after the operands • (4+3)*5 : 4 3 + 5 * • 4+(3*5) : 4 3 5 * + – No precedence rules or parentheses! • Input expression given in postfix form – How to evaluate it?
- 3. Evaluating Postfix Expressions • Use a stack, assume binary operators +,* • Input: postfix expression • Scan the input – If operand, • push to stack – If operator • pop the stack twice • apply operator • push result back to stack
- 4. Example • Input 5 9 8 + 4 6 * * 7 + * • Evaluation push(5) push(9) push(8) push(pop() + pop()) /* be careful for ‘-’ */ push(4) push(6) push(pop() * pop()) push(7) push(pop() + pop()) push(pop() * pop()) print(pop()) • What is the answer?
- 5. Exercise • Input 6 5 2 3 + 8 * + 3 + * • Input a b c * + d e * f + g * + • For each of the previous inputs – Find the infix expression
- 6. Infix to Postfix Conversion • Observation – Operands appear in the same order in both – Output operands as we scan the input – Must put operators somewhere • use a stack to hold pending operators – ‘)’ indicates both operands have been seen • Will allow only +, *, ‘(‘, and ‘)’, and use standard precedence rules • Assume legal (valid) expression
- 7. Conversion Algorithm • Output operands as encountered • Stack left parentheses • When ‘)’ – repeat • pop stack, output symbol – until ‘(‘ • ‘(‘ is poped but not output • If symbol +, *, or ‘(‘ – pop stack until entry of lower priority or ‘(‘ • ‘(‘ removed only when matching ‘)’ is processed – push symbol into stack • At end of input, pop stack until empty
- 8. Exercises • (5 * (((9 + 8) * (4 * 6)) + 7)) • 6 * (5 + (2 + 3) * 8 + 3) • a + b * c + (d * e + f) * g