how can i convert infix to postfix expression of multi digit operands? - c++

Question

I am able to convert infix to postfix and calculates with one digit but i can't convert to postfix and calculate with N digits. PLz anyone help me! Thanks!!
here is my code with single digit
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
class infix2postfix
{
public:
void push(int symbol);
int pop();
void infix_to_postfix();
int priority(char symbol);
int isEmpty();
int white_space(char);
int eval_post();
};
char infix[100], postfix[100];
int stack[100];
int top;
int main()
{
infix2postfix ip;
top=-1;
cout<<"Enter infix : ";
gets(infix);
ip.infix_to_postfix();
cout<<"Postfix : "<<postfix<<endl;
cout<<"Result is : "<<ip.eval_post()<<endl;
return 1;
}
void infix2postfix :: infix_to_postfix()
{
int i,p=0;
char next;
char symbol;
for(i=0; i<strlen(infix); i++)
{
symbol=infix[i];
if(!white_space(symbol))
{
switch(symbol)
{
case '(':
push(symbol);
break;
case ')':
while((next=pop())!='(')
postfix[p++] = next;
break;
case '+':
case '-':
case '*':
case '/':
case '%':
case '^':
while( !isEmpty( ) && priority(stack[top])>= priority(symbol) )
postfix[p++]=pop();
push(symbol);
break;
default: /*if an operand comes*/
postfix[p++]=symbol;
}
}
}
while(!isEmpty( ))
postfix[p++]=pop();
postfix[p]='\0'; /*End postfix with'\0' to make it a string*/
}
/*This function returns the priority of the operator*/
int infix2postfix :: priority(char symbol)
{
switch(symbol)
{
case '(':
return 0;
case '+':
case '-':
return 1;
case '*':
case '/':
case '%':
return 2;
case '^':
return 3;
default :
return 0;
}
}
void infix2postfix :: push(int symbol)
{
if(top>100)
{
cout<<"Stack overflow\n";
exit(1);
}
stack[++top]=symbol;
}
int infix2postfix :: pop()
{
if( isEmpty() )
{
cout<<"Stack underflow\n";
exit(1);
}
return (stack[top--]);
}
int infix2postfix :: isEmpty()
{
if(top==-1)
return 1;
else
return 0;
}
int infix2postfix :: white_space(char symbol)
{
if( symbol == ' ' || symbol == '\t' )
return 1;
else
return 0;
}
int infix2postfix :: eval_post()
{
int a,b,i,temp,result;
for(i=0; i<strlen(postfix); i++)
{
if(postfix[i]<='9' && postfix[i]>='0')
push(postfix[i]-'0');
else
{
a=pop();
b=pop();
switch(postfix[i])
{
case '+':
temp=b+a;
break;
case '-':
temp=b-a;
break;
case '*':
temp=b*a;
break;
case '/':
temp=b/a;
break;
case '%':
temp=b%a;
break;
case '^':
temp=pow(b,a);
}
push(temp);
}
}
result=pop();
return result;
}
My Question is: how to get the result for more than 1 digit operand?
I have tried for Single digit, but how could i get for multi-digit numbers?

Answer 1

You are currently pushing the digits on the stack individually, so a numeric value of 10 will be pushed on the stack two symbols: 1 and 0.
In your operator logic, you are popping one symbol from the stack per operand. Multi-digits operands will thus not work and produce completely wrong results.
There are many ways to solve this. For example, you could fuse multiple digits together into the same stack symbol while reading them in (i.e. combine the digit currently processed with the top of the stack if both are digits).
The place to do this would be inside push. Here's how to do this:
void infix2postfix :: push(int symbol)
{
if(top>100)
{
cout<<"Stack overflow\n";
exit(1);
}
if (! isEmpty() && stack[top] >= 0 && stack[top] <= 9) {
stack[top] *= 10;
stack[top] += symbol;
}
else {
stack[++top]=symbol;
}
}
This will work as long as a numeric operand fits inside the range of an int. Bigger numbers will simply overflow.
There will also be problems because numbers cannot be told apart from other stack symbols. With multi-digit numbers, you can now have symbols which have the same int value as an operator. A way to fix this is to assign negative int values to the operator symbols, and non-negative values for the numbers. This will work for your case as your grammar does not seem to have a unary minus.
This approach is also in line with what you do in eval_post, where you push the calculation result on the stack as a single integer value, regardless of how many digits might be contained.
The most powerful but complicated alternative would write a grammar and use a parser generator. I recommend GNU bison. This will completely take over generating the code for the parser, so all you have to do is to write the grammar for your expressions plus the actual infix-to-postfix transformation (which will be trivial with bison). It will also help you detect invalid input easily and provide appropriate error messages.
However, getting started with bison can be hard if you never used it before.