Evaluate Infix expression without converting it into postfix [duplicate] - c++

Question

This question already has answers here:
How to evaluate an infix expression in just one scan using stacks?
(4 answers)
Closed 3 years ago.
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;
}

Answer 1

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 )