Reverse Polish notation Calculator - c++

Question

I am currently working on a RPN calculator, it takes an infix expression converts it to postfix and shows the answer. I mostly got it right, but when I pop the answer from the stack if shows only the last digit of the result
ex
Enter infix: (1+1)*13+10/2
Postfix: 11+13*102/+
Result is: 1
Enter infix: 2*13+10/2
Postfix: 213*102/+
Result is:1
It gets it right for this kind of inputs
Enter infix: 3*2+5
Postfix: 32*5+
Result is : 11
#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;
}

Answer 1

Consider what happens when eval_post() is given 213*102/+ to work with. Let's start in the middle, with the '1' after the asterisk. The '1' is a digit, so push it [stack ends with: 1]. Similarly, the 0 and 2 get pushed [stack ends with: 1, 0, 2]. Then the division symbol is encountered, so pop 2 and 0, then push 0/2 = 0 [stack ends with: 1, 0]. Finally, the addition symbol is encountered, so pop 0 and 1, then push 1+0=1, which is then popped as your answer.
One symptom of your problem is that, if things work, the stack should be empty when eval_post() returns. However, it is not empty when your infix includes numbers with more than one digit. Note that "10" gets pushed onto the stack as two numbers: "1" followed by "0". You want the value 10 pushed.
There are also some style problems with the code, but this appears to be the main functional problem.