Evaluate Infix expression without converting it into postfix

Question

I'm trying to evaluate an infix expression in 1 pass without converting it into postfix but it's not giving correct output for some expressions. For eg: 3-5*10/5+10 , (45+5)-5*(100/10)+5

Can someone provide a proper solution to this problem in cpp.

Link to the previous question asked: How to evaluate an infix expression in just one scan using stacks?

Please don't mark it as duplicate as I have tried the algorithm answered in the above given thread but to no avail.

#include<bits/stdc++.h>

int isoperand(char x)
{
    if(x == '+' || x=='-'|| x=='*' || x=='/' || x==')' || x=='(')
        return 0;
    return 1;
}

int Pre(char x)
{
    if(x == '+' || x == '-')
        return 1;
    if(x == '*' || x == '/')
        return 3;
    return 0;
}

int infixevaluation(std::string exp)
{
    std::stack<int> s1; //Operand Stack
    std::stack<char> s2; //Operator Stack
    int i,x,y,z,key;
    i=0;
    while(exp[i]!='\0')
    {

        if(isoperand(exp[i]))
        {
            key = exp[i]-'0';
            s1.push(key);
            i++;
        }
        else if(!isoperand(exp[i]) && s2.empty())
            s2.push(exp[i++]);
        else if(!isoperand(exp[i]) && !s2.empty())
        {
            if(Pre(exp[i])>Pre(s2.top()) && exp[i]!=')')
                s2.push(exp[i++]);
            else if(exp[i]==')' && s2.top() == '(')
            {
                s2.pop();
                i++;
            }
            else if(exp[i]=='(')
                s2.push(exp[i++]);
            else
            {
                x = s1.top();
                s1.pop();
                y = s2.top();
                s2.pop();
                z = s1.top();
                s1.pop();
                if(y == '+')
                    s1.push(z+x);
                else if(y == '-')
                    s1.push(z-x);
                else if(y == '*')
                    s1.push(x*z);
                else if(y == '/')
                    s1.push(z/x);
            } 
        }
    }
    while(!s2.empty())
    {
        x = s1.top();
        s1.pop();
        y = s2.top();
        s2.pop();
        z = s1.top();
        s1.pop();
        if(y == '+')
            s1.push(x+z);
        else if(y == '-')
            s1.push(z-x);
        else if(y == '*')
            s1.push(x*z);
        else if(y == '/')
            s1.push(z/x);
    }
    return s1.top();
}

int main(int argc, char const *argv[])
{
    std::string s;
    getline(std::cin,s);
    std::cout<<infixevaluation(s)<<std::endl;
    return 0;
}

What is the expected output, and what is the actual output? Also, does [debugging your program](http://ericlippert.com/2014/03/05/how-to-debug-small-programs/) allow you to narrow down where the problem lies? — JaMiT, Aug 22 '19 at 02:30
It's giving output as floating-point exception for the expression (45+5)-5*(100/10)+5 and for 3-5*10/5+10 output is 1 instead of 3. — Amey, Aug 22 '19 at 02:35
An exception is a crash, not output. Your next step should be to [debug your program](http://ericlippert.com/2014/03/05/how-to-debug-small-programs/) to determine where the crash occurs. Then simplify your program (removing functionality) until you have a [mcve] demonstrating the crash. — JaMiT, Aug 22 '19 at 02:50
Be careful with `#include` It includes pretty much the entire Standard library, turning your code into something of a minefield . You haven't tacked on `using namespace std;`, which really makes stdc++.h a bad idea by turning your code into a really big minefield, but there are a [whole bunch of other reasons not to use it.](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — user4581301, Aug 22 '19 at 03:53
Thanks for asking this question by the way. Including stdc++.h when I took a look at your code found that something corrupted the copy of shared_ptr_base.h in my gcc install. Good to find these things when it's not at a bad time. Think I should consider getting a new hard drive too. Sucker's probably 9 years old. — user4581301, Aug 22 '19 at 04:13
Your code would be a whole lot easier to understand if you used meaningful variable names. For example, `s1` could be `operandStack`. And instead of `x = s1.top()`, how about `operand1 = operandStack.top()`? In a production environment, I wouldn't even check this code for proper operation until it had meaningful variable names. — Jim Mischel, Aug 22 '19 at 22:44
@user4581301 hahaha... thank you for giving me advice about not including that particular header file. — Amey, Aug 22 '19 at 22:50
'To no avail' is not a problem description. The code in the duplicate is correct. — user207421, Aug 22 '19 at 23:49

score 1 · Accepted Answer · answered Aug 22 '19 at 14:16

Your code can only deal with single-digit operands -- and it has no checks for malformed input, so when you have a multi-digit operand, it runs off the rails.

The easiest fix for the former is just to scan digits when you see a digit -- change the if (isoperand(exp[i]) clause to:

    if (isdigit(exp[i])) {
        int value = 0;
        while (isdigit(exp[i]))
            value = value * 10 + exp[i++] - '0';
        s1.push(value);
    } else ...

For error checking, you should do things like

check for spaces and other invalid characters and reject or skip them
keep track of whether the last token matched was an operand or an operator, and give errors for two consecutive operands, or two consecutive operators other than ( and )

Thank you! It worked. I was using pen and paper to trace the code and I was pushing 2 digit number at once. Should have used a debugger. Silly me! Link to the full solution: http://ideone.com/IV8PAU — Amey, Aug 22 '19 at 23:34

Evaluate Infix expression without converting it into postfix

1 Answers1