Member Avatar for Ineedhelpplz

So I have written a reverse polish notation calculator:

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>

#define NUMBER '0'
#define MAXOP   100

int main ()
{
        int type;
        int op1, op2;
        char    s[MAXOP];
        while ((type = getop(s)) != EOF)
        switch (type)
        {
        case NUMBER:
          push(atoi(s));
          break;
        case '^':
          push ((unsigned int) pop() ^ (unsigned int) pop());
          break;
        case '~':
          push(~(unsigned int) pop());
          break;
        case '+':
        case '*':
          if (type == '+')
          push(pop() + pop());
          else
          push(pop() * pop());
          break;
        case '-':
          op2 = pop();
          push(pop() - op2);
          break;
        case '/':
        case '%':
          if ((op2 = pop()) != 0.0) {
                if (type == '/')
                        push(pop() / op2);
                else {
                        op1 = pop();
                        push(op1 - op2 * ((int) (op1/op2)));
                     }
        } else
                printf("Error: Zero divisor!\n");
                break;
        case '\n':
                printf("The answer is %d\n", pop());
                break;
        default:
                printf("Error: Unknown command %s!\n", s);
                break;
        }
                return 0;
}

I now need to run the infix string " ~(((202%16) + (292/16)*16) ^292 " through my calculator...however the string is in infix, RPN takes postfix. I have been having trouble converting this expression to postfix and need help. The program itself is done I just want to test that input and don't quite understand how to make it work in postfix.

Edited by Ineedhelpplz because: fixed code
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Dani AI

Generated

Quick, practical notes that build on 's pointer to existing threads and 's example.

The important semantics: ^ in your calculator is bitwise XOR (not exponentiation) and ~ is a unary prefix bitwise-NOT. In postfix the prefix unary operator becomes a postfix token (so ~x turns into x ~). Precedence you should enforce (highest → lowest): unary ~, multiplicative * / %, additive + -, then bitwise XOR ^. Treat unary ~ as right-associative; all the binary operators used here are left-associative.

A concise shunting-yard outline to convert infix → postfix:

  • Tokenize numbers, operators and parentheses. Ensure multi-digit numbers are single tokens.
  • If token is a number: output it.
  • If token is an operator O:
    • while stack top is operator with greater precedence, or equal precedence and left-associative, pop it to output
    • push O
  • If token is (: push it. If ): pop operators to output until ( is popped.
  • After input ends, pop remaining operators to output.

Handle unary operators explicitly while tokenizing: when an operator appears at expression start or immediately after ( or another operator, treat ~ as unary. Represent it internally (e.g., u~) so it gets the correct precedence and is emitted after its operand.

Practical tips for your existing calculator:

  • Feed RPN as space-separated tokens so getop/atoi reads numbers correctly.
  • Verify integer semantics: division and modulo are integer operations; negative operands follow C rules. Use op2 != 0 (integer 0) for divisor checks to avoid mixing float literals.
  • If results are wrong, test and convert small subexpressions first, confirm parentheses balance, and confirm your converter distinguishes unary vs binary operators.

Recommended Answers

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Member Avatar for dkalita

Just explore this site a bit.
Infix to postfix conversion has been discussed earlier in many threads.

Member Avatar for Ineedhelpplz

Thanks! Got my answer:

292 16 % 292 16 / + 16 * 292 ^ ~

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