{"id":986,"date":"2025-11-21T09:01:19","date_gmt":"2025-11-21T03:31:19","guid":{"rendered":"https:\/\/codexplained.in\/?p=986"},"modified":"2025-11-21T09:01:19","modified_gmt":"2025-11-21T03:31:19","slug":"evaluation-of-postfix-expression","status":"publish","type":"post","link":"https:\/\/codexplained.in\/?p=986","title":{"rendered":"Evaluation of Postfix Expression"},"content":{"rendered":"\n

How to Evaluate a Postfix Expression<\/h3>\n\n\n\n
    \n
  1. Traverse the expression<\/strong>: We iterate through each character of the postfix expression from left to right.<\/li>\n\n\n\n
  2. Push operands onto a stack<\/strong>: When encountering a number (operand), push it onto a stack.<\/li>\n\n\n\n
  3. Apply operators<\/strong>: When an operator is encountered, pop the required number of operands from the stack, perform the operation, and push the result back onto the stack.<\/li>\n\n\n\n
  4. End condition<\/strong>: After traversing the entire expression, the result will be the single value left on the stack.<\/li>\n<\/ol>\n\n\n\n

    Algorithm<\/h3>\n\n\n\n
      \n
    1. Initialize an empty stack.<\/li>\n\n\n\n
    2. Read the postfix expression character by character:\n
        \n
      • If the character is an operand, push it to the stack.<\/li>\n\n\n\n
      • If the character is an operator, pop two operands from the stack, apply the operator, and push the result back onto the stack.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • At the end, the value left in the stack is the result of the postfix expression.<\/li>\n<\/ol>\n\n\n\n

        Example<\/h3>\n\n\n\n

        Let’s take the postfix expression:<\/p>\n\n\n

        \n53+82-*\n<\/pre><\/div>\n\n\n
        This corresponds to the infix expression:<\/p>\n\n\n
        \n(5 + 3) * (8 - 2) = 48\n<\/pre><\/div>\n\n\n
        C Program to Evaluate Postfix Expression<\/p>\n\n\n
        \n#include <stdio.h>\n#include <ctype.h>  \/\/ For isdigit() function\n#include <stdlib.h> \/\/ For exit() function\n\n#define MAX 100\n\n\/\/ Stack structure\nint stack[MAX];\nint top = -1;\n\n\/\/ Push function\nvoid push(int value) \n{\n    if (top == MAX - 1) \n{\n        printf("Stack Overflow\\n");\n        exit(1);\n    }\n    stack[++top] = value;\n}\n\n\/\/ Pop function\nint pop() \n{\n    if (top == -1) \n{\n        printf("Stack Underflow\\n");\n        exit(1);\n    }\n    return stack[top--];\n}\n\n\/\/ Function to evaluate postfix expression\nint evaluatePostfix(char* expression) \n{\n    int i = 0;\n    while (expression[i] != '\\0') \n{\n        \/\/ If the character is an operand (digit), push it to the stack\n        if (isdigit(expression[i])) \n{\n            push(expression[i] - '0');  \/\/ Convert char to int\n        } \n        \/\/ If the character is an operator, pop two elements and apply the operator\n        else \n{\n            int val1 = pop();\n            int val2 = pop();\n            \n            switch (expression[i]) \n{\n                case '+': push(val2 + val1); break;\n                case '-': push(val2 - val1); break;\n                case '*': push(val2 * val1); break;\n                case '\/': push(val2 \/ val1); break;\n                default: \n                    printf("Invalid operator encountered\\n");\n                    exit(1);\n            }\n        }\n        i++;\n    }\n    \n    \/\/ The result will be the last element in the stack\n    return pop();\n}\n\nint main() \n{\n    char postfixExpression[] = "53+82-*";\n    printf("Postfix Expression: %s\\n", postfixExpression);\n    int result = evaluatePostfix(postfixExpression);\n    printf("Result of the Postfix Expression: %d\\n", result);\n    return 0;\n}\n<\/pre><\/div>\n\n\n
        Explanation of the Code<\/h3>\n\n\n\n
          \n
        1. Stack<\/strong>: We use an array stack[]<\/code> to implement the stack, with top<\/code> keeping track of the top of the stack.<\/li>\n\n\n\n
        2. Push and Pop Functions<\/strong>: These handle adding and removing elements from the stack.\n
            \n
          • push(int value)<\/code> adds an element to the top of the stack.<\/li>\n\n\n\n
          • pop()<\/code> removes the top element from the stack and returns it.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Main Function<\/strong>:\n
              \n
            • The postfix expression 53+82-*<\/code> is passed to the evaluatePostfix()<\/code> function.<\/li>\n\n\n\n
            • The function processes each character of the expression. If it is a digit, it is pushed onto the stack. If it is an operator, two operands are popped from the stack, the operation is performed, and the result is pushed back onto the stack.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Switch-Case<\/strong>: Used to apply the correct operator on the two popped values.<\/li>\n\n\n\n
            • Final Result<\/strong>: After evaluating the entire expression, the result is stored at the top of the stack, and we print it.<\/li>\n<\/ol>\n\n\n\n
              Example Walkthrough<\/h3>\n\n\n\n
              For the expression 53+82-*<\/code>:<\/p>\n\n\n\n
                \n
              1. Push 5<\/code> onto the stack.<\/li>\n\n\n\n
              2. Push 3<\/code> onto the stack.<\/li>\n\n\n\n
              3. Encounter +<\/code>: Pop 5<\/code> and 3<\/code>, calculate 5 + 3 = 8<\/code>, push 8<\/code>.<\/li>\n\n\n\n
              4. Push 8<\/code> onto the stack.<\/li>\n\n\n\n
              5. Push 2<\/code> onto the stack.<\/li>\n\n\n\n
              6. Encounter -<\/code>: Pop 8<\/code> and 2<\/code>, calculate 8 - 2 = 6<\/code>, push 6<\/code>.<\/li>\n\n\n\n
              7. Encounter *<\/code>: Pop 8<\/code> and 6<\/code>, calculate 8 * 6 = 48<\/code>, push 48<\/code>.<\/li>\n<\/ol>\n\n\n\n
                Final result: 48<\/code>.<\/p>\n\n\n\n
                Output<\/strong><\/p>\n\n\n
                \nPostfix Expression: 53+82-*\nResult of the Postfix Expression: 48\n<\/pre><\/div>\n\n\n
                This simple program demonstrates how to evaluate postfix expressions using a stack. By using stack operations (push and pop), we can easily handle any postfix expression.<\/p>\n\n\n\n
                <\/p>\n