How to evaluate a postfix expression using character stack using ASCII conversions

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The code is implemented using ADT by including the "stack.h" user defined header file.

The header file has code for stack operation. I am getting wrong outputs for results greater than 9.

What changes should I make in the code to get the correct output.

I have to use a stack with char data and not int data.

#include<iostream>
#include<string.h>
#include "stack.h"
using namespace std;

void posteva(char postfix[])
{
    stack s;
    int i=0;
    while(postfix[i]!='\0')
    {
        char x=postfix[i];
        if(isdigit(x))
        {
            s.push(x);
        }
        else
        {
            int op1=s.pop()-'0';
            int op2=s.pop()-'0';
            int res;
            switch(x)
            {
                case '+':
                    res=op1+op2;
                    break;
                case '-':
                    res=op1-op2;
                    break;
                case '*':
                    res=op1*op2;
                    break;
                case '/':
                    res=op1/op2;
                    break;
                case '%':
                    res=op1%op2;
                    break;

            }
            s.push(res+'0');
        }
        i++;
    }
    cout<<"\n\nRESULT :"<<s.pop();
}

int main()
{
    char postfix[20];
    cout<<"\n\nEnter the postfix : ";
    cin>>postfix;
    posteva(postfix);
}

What I have tried:

For example, for postfix expression "63*" I am getting result as B.

Posted 18-Apr-18 2:59am

NiRaj Wagh

Updated 18-Apr-18 4:06am

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Richard MacCutchan · Answer 1 · 2018-04-18T03:19:00

Solution 1

C++

s.push(res+'0');

That will only work for a single digit result. In your case, 6 * 3 = 18. The character '0' is equivalent to 0x30, or decimal 48. 48 + 18 = 66 or hexadecimal 0x42, which is the numeric value of the ASCII character 'B'.

Posted 18-Apr-18 3:19am

Richard MacCutchan

Patrice T · Answer 2 · 2018-04-18T03:28:00

Quote:
for postfix expression "63*" I am getting result as B.

Guilty is:

C++

s.push(res+'0');

because the code is correct only for single digit values.

In order to handle numbers with more than 1 digit, you need to have separator between numbers, a space is usually used for this:

C++

2 63 3*+

My advice is to push values on stack after converting to integers, but you have to rewrite your code to handle entry of number > 9.

Jochen Arndt · Answer 3 · 2018-04-18T04:07:00

You are pushing single digits / characters on the stack. When a number is greater than 9 you will push multiple digits that must be all popped and converted to a number.

It all depends on how the input is formatted. Assuming something like "num1 num2 op", I suggest to write functions to push and pop numbers:

C++

int pushdigit(const char* input, int pos, stack *s)
{
    while (istdigit(input[pos])
    {
        s->push(input[pos++]);
    }
    // Push a zero as end of number indicator
    s->push(0);
    // Return position to next input character skipping space
    return (' ' == input[pos]) ? pos + 1 : pos;
}

int popdigit(stack *s)
{
    int num = 0;
    int mult = 1;
    // Assuming the stack provides a function to check if it is empty
    while (!s->isempty())
    {
        char c = s->pop();
        if (!isdigit(c))
            break;
        num += mult * (c - '0');
        mult *= 10;
    }
    return num;
}

How to evaluate a postfix expression using character stack using ASCII conversions

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