Infix to Postfix Example
Pre-requisite :
| Expression | Current Symbol | Stack | Output | Comment |
| A/B^C-D | Initial State | NULL | - | Initially Stack is Empty |
| /B^C-D | A | NULL | A | Print Operand |
| B^C-D | / | / | A | Push Operator Onto Stack |
| ^C-D | B | / | AB | Print Operand |
| C-D | ^ | /^ | AB | Push Operator Onto Stack because Priority of ^ is greater than Current Topmost Symbol of Stack i.e ‘/’ |
| -D | C | /^ | ABC | Print Operand |
| D | - | / | ABC^ | Step 1 : Now ‘^’ Has Higher Priority than Incoming Operator So We have to Pop Topmost Element . Step 2 : Remove Topmost Operator From Stack and Print it |
| D | - | NULL | ABC^/ | Step 1 : Now ‘/’ is topmost Element of Stack Has Higher Priority than Incoming Operator So We have to Pop Topmost Element again. Step 2 : Remove Topmost Operator From Stack and Print it |
| D | - | - | ABC^/ | Step 1 : Now Stack Becomes Empty and We can Push Operand Onto Stack |
| NULL | D | - | ABC^/D | Print Operand |
| NULL | NULL | - | ABC^/D- | Expression Scanning Ends but we have still one more element in stack so pop it and display it |
