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# 1  
Old 03-15-2011
An error in my c++ program... please help urgent.
The following is a program to convert an infix expression to postfix expression.


//Convert an infix expression to postfix expression...

Code:
#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
char ifx[50],pfx[50],stk[50];
int top=-1,n;
void push(char ch)
{
    if(top!=n)
    {
        ch=stk[top];
        top++;
        cout<<"\nElement added was "<<ch;
    }
    else
        cout<<"\nThe stack is full.";    
}
void pop()
{
    char rmv;
    if(top!=-1)
    {
        rmv=stk[top];
        top--;
        cout<<"\nElement removed was "<<rmv;
    }
    else
        cout<<"\nThe stack is empty.";
}
char topele()
{
    char t;
    if(top!=-1)
        t=stk[top];
    else
    {
        t='#';
        cout<<"\nThere are no elements in the stack.The stack is: "<<t;
    }
    return t;
}
int chkpres()
{
    char ch;
    switch(ch)
    {
        case '^': return 5;
            break;
        case '*': return 4;
            break;
        case '/': return 3;
            break;
        case '+': return 2;
            break;
        case '-': return 1;
            break;
        default: exit(1);
            break;
    }
}
int main()
{
    char pre,pres,ele,elem,chk,popp,topp;
    cout<<"\nEnter how many elements you want to enter in the infix expression: ";
    cin>>n;
    for(int j=0;j<n;++j)
    {
        cout<<"\nEnter the characters of the infix expression one by one: ";
        cin>>ifx[j];
    }
    topp=topele();
    for(int i=0;ifx[i]='\0';++i)
    {
        if(ifx[i]=='(')
        {
            break;
        }
        if(ifx[i]=='^'||ifx[i]=='*'||ifx[i]=='/'||ifx=='+'||ifx[i]=='-')
        {
            if(topp=='^'||topp=='*'||topp=='/'||topp=='+'||topp=='-')
            {
                pre=ifx[i];
                pres=topp;
                chk=chkpres();
                if(pre>pres)
                {
                    topp=pre;
                    top++;
                }
                else if(pres>=pre)
                    pfx[i]=pres;
            }    
        }
        else if(ifx[i]=')')
        {
            while(topp!='#')
            {
                popp=topp;
                pfx[i]=popp;
            }
        } 
    }
    cout<<"\nThe postfix expression is: ";
    for(int i=0;pfx[i]!='\0';++i)
    {
        cout<<pfx[i];
    }
return 0;
}

Here's the output of the above program :


Code:
lab1.cpp:80: error: ISO C++ forbids comparison between pointer and integer


Please suggest me to resolve this problem...
Last edited by pludi; 03-16-2011 at 05:04 AM.. Reason: code tags, please
# 2  
Old 03-15-2011
Use code tags to post code. Not using them destroyed your formatting.

taking a look.

---------- Post updated at 09:44 AM ---------- Previous update was at 09:43 AM ----------

You forgot to dereference your array in one place: ifx=='+'
This User Gave Thanks to Corona688 For This Post:
poonam.gaigole
# 3  
Old 03-16-2011
Oh... thank you so much... I should have been able to notice that... Well, thank you again.

---------- Post updated 03-16-11 at 10:42 AM ---------- Previous update was 03-15-11 at 09:19 PM ----------

The program is:



Code:
//Convert an infix expression to postfix expression...

#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
char ifx[50],pfx[50],stk[50];
int top=-1,n;
char push(char ch)
{
    if(top!=n)
    {
        ch=stk[top];
        top++;
        return ch;
    }
    else
        cout<<"\nThe stack is full.";    
}
void pop()
{
    char rmv;
    if(top!=-1)
    {
        rmv=stk[top];
        top--;
        cout<<"\nElement removed was "<<rmv;
    }
    else
        cout<<"\nThe stack is empty.";
}
char topele()
{
    char t;
    if(top==-1)
        t='#';
    else
        t=stk[top];
    return t;
}
int chkpres()
{
    char ch;
    switch(ch)
    {
        case '^': return 5;
            break;
        case '*': return 4;
            break;
        case '/': return 3;
            break;
        case '+': return 2;
            break;
        case '-': return 1;
            break;
        default: exit(1);
            break;
    }
}
int main()
{
    char pre,pres,ele,elem,chk,popp,topp;
    cout<<"\nEnter how many elements you want to enter in the infix expression: ";
    cin>>n;
    topp=stk[top];
    cout<<"\nEnter the characters of the infix expression one by one: ";
    for(int i=0;i<n;++i)
    {
        cin>>ifx[i];
        //for(int i=0;ifx[i]='\0';++i)
        //{
            if(ifx[i]!='^' && ifx[i]!='*' && ifx[i]!='/' && ifx[i]!='+' && ifx[i]!='-')
                pfx[i]=ifx[i];
            else if(ele=='(')
            {
                ele=ifx[i];
                push(ele);
                top++;
            }
            else if(ifx[i]=='^'||ifx[i]=='*'||ifx[i]=='/'||ifx[i]=='+'||ifx[i]=='-')
            {
                if(topp=='^'||topp=='*'||topp=='/'||topp=='+'||topp=='-')
                {
                    pre=ele;
                    pres=topp;
                    chk=chkpres();
                    if(pre>pres)
                    {
                        topp=pre;
                        top++;
                    }
                    else if(pres>=pre)
                        pfx[i]=pres;
                }    
            }
            else if(ele=')')
            {
                while(topp!='#')
                {
                    popp=topp;
                    pfx[i]=popp;
                }
            }
       // }
    }
    cout<<"\nThe postfix expression is: ";
    for(int i=0;pfx[i]!='\0';++i)
    {
        cout<<pfx[i];
    }
return 0;
}




OUTPUT is:

Code:
Enter how many elements you want to enter in the infix expression: 9

Enter the characters of the infix expression one by one: a
*
(
b
+
c
)
/
d

The postfix expression is: a




Could you please help me to fix it...
Last edited by pludi; 03-16-2011 at 05:05 AM.. Reason: code tags, please
# 4  
Old 03-16-2011
The code you gave didn't even work as you showed it, it went into an infinite loop...

I've tried to clean it up a little, but there's problems I don't know how to fix since I don't know what you were actually trying to do:

Code:
//Convert an infix expression to postfix expression...

#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
char ifx[50],pfx[50],stk[50];
int top=-1,n;
char push(char ch)
{
    if(top!=n)
    {
        ch=stk[top];
        top++;
        return ch;
    }
    else
        cout<<"\nThe stack is full.";
}
void pop()
{
    char rmv;
    if(top!=-1)
    {
        rmv=stk[top];
        top--;
        cout<<"\nElement removed was "<<rmv;
    }
    else
        cout<<"\nThe stack is empty.";
}
char topele()
{
    char t;
    if(top==-1)
        t='#';
    else
        t=stk[top];
    return t;
}
int chkpres()
{
    char ch;
    switch(ch)
    {
        case '^': return 5;
            break;
        case '*': return 4;
            break;
        case '/': return 3;
            break;
        case '+': return 2;
            break;
        case '-': return 1;
            break;
        default: exit(1);
            break;
    }
}
int main()
{
    int i;

    char pre,pres,ele,elem,chk,popp,topp;
    cout<<"\nEnter how many elements you want to enter in the infix expression: ";
    cin>>n;
    topp=stk[top];
    cout<<"\nEnter the characters of the infix expression one by one: ";

    // Make this its own loop to simplify the rest
    for(i=0;i<n;++i)
    {
        cin>>ifx[i];
    }
    // You can't assume the array will end in '\0', you have to put it there
    ifx[i]='\0';

    // uncommented this.  It was a good idea, but didn't 
    // work before because you forgot the !
    for(i=0;ifx[i]!='\0';++i)
    {
            // Is this really what you want it to do?  Nothing except letters
            // and brackets will make to the second if-statement.
            if(ifx[i]!='^' && ifx[i]!='*' && ifx[i]!='/' && ifx[i]!='+' && ifx[i]!='-')
                pfx[i]=ifx[i];
            // 'ele' may not even have been set yet, in which case its value will
            // be undefined.  Which could be (.  Or could be \xfe.  What do you
            // want 'ele' to be by default?  set it when you create the var.
            else if(ele=='(')
            {
                ele=ifx[i];
                push(ele);
                top++;
                // Shouldn't you set pfx[i] to something here?  Otherwise it
                // will stay at an undefined value at index i -- possibly \0,
                // possibly garbage
            }
            else if(ifx[i]=='^'||ifx[i]=='*'||ifx[i]=='/'||ifx[i]=='+'||ifx[i]=='-')
            {
                if(topp=='^'||topp=='*'||topp=='/'||topp=='+'||topp=='-')
                {
                    pre=ele;
                    pres=topp;
                    chk=chkpres();
                    if(pre>pres)
                    {
                        topp=pre;
                        top++;
                        // Shouldn't you set pfx[i] to something here?  Otherwise it
                        // will stay at an undefined value at index i -- possibly \0,
                        // possibly garbage
                    }
                    else if(pres>=pre)
                        pfx[i]=pres;
                }
            }
            // There is a huge difference between = and ==.
            // you were changing ele to ')' here.
            // 'ele' may not even have been set yet, either, see above.
            else if(ele==')')
            {
                while(topp!='#')
                {
                    popp=topp;
                    pfx[i]=popp;
                }
            }
       // }
    }

    // You can't assume the array will end in '\0', you have to put it there
    pfx[i]='\0';

    cout<<"\nThe postfix expression is: ";
    for(i=0; pfx[i]!='\0'; ++i)
    {
        cout<<pfx[i];
    }
return 0;
}

So I can't fix it outright but I hope I've given you a better idea of what's going wrong.
This User Gave Thanks to Corona688 For This Post:
poonam.gaigole
# 5  
Old 03-17-2011
Why cannot I assume in my program the arrays ifx[] and pfx[] to end at '\0' ?
Because since these are string arrays, I think we should assume it to end with a null character.
Why we have to specify/initialize that the two arrays would end at null character?
# 6  
Old 03-17-2011
Quote:
Originally Posted by poonam.gaigole
Why cannot I assume in my program the arrays ifx[] and pfx[] to end at '\0' ?
Because any contents you don't initialize yourself are undefined.
Quote:
Because since these are string arrays, I think we should assume it to end with a null character.
You have it exactly backwards. A string doesn't define a NULL -- a NULL defines a string. You've got chars arrays, which are just integers but smaller and have no mysterious auto-setting properties. Would you expect setting a[0] to also set a[1] for int? You were just lucky enough to be using global vars that "usually" default to 0. Try that with stack variables and you're in for a big surprise.
Quote:
Why we have to specify/initialize that the two arrays would end at null character?
If not you, then who?
Last edited by Corona688; 03-17-2011 at 03:23 AM..
# 7  
Old 03-17-2011
Here, 'ele' is the element that we read from the infix expression (ifx[]) and I think i should not have used the 'elem' because its of no use in my program.
The following code compares between the elements present in the two stacks one by one.


Code:
Code:
        else if(ifx[i]=='^'||ifx[i]=='*'||ifx[i]=='/'||ifx[i]=='+'||ifx[i]=='-')
            {
                 if(topp=='^'||topp=='*'||topp=='/'||topp=='+'||topp=='-')
                {
                    pre=ele;
                    // 'pre' is the current (i) element in the infix expression.
         
                    pres=topp;
                    // 'pres' is the topmost element of the stack. 
                    chk=chkpres();
                    // Here I'm checking the precedence of the operators.

                    if(pre>pres)
                    {
                        topp=pre;
                        top++;
                    }
                    else if(pres>=pre)
                         pfx[i]=pres;
                }    
                else
                {
                    ele=ifx[i];
                    push(ele);
                }
                ++i;
            }



Well, I'm trying my level best to solve this hoping to come out with the appropriate result. Thank you very much for your patience.
Last edited by pludi; 03-17-2011 at 08:18 AM..
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