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-   -   InFix to PostFix and PostFix expression evaluation. (http://www.go4expert.com/articles/infix-postfix-postfix-expression-t1693/)

shabbir 21Oct2006 10:30

InFix to PostFix and PostFix expression evaluation.
 
1 Attachment(s)
InFix to PostFix

Introduction



Infix Expression :

Notation in which the operator separates its operands. Eg (a + b) * c. Infix notation requires the use of brackets to specify the order of evaluation.

Postfix Expression :

Reverse Polish Notation or Suffix Notation Notation in which the operator follows its operands. Eg a + b * c represented as abc*+.

Infix to Postfix Conversion Algo :
  1. Scan the Infix string from left to right.
  2. Initialise an empty stack.
  3. If the scannned character is an operand, add it to the Postfix string. If the scanned character is an operator and if the stack is empty Push the character to stack.
  4. If the scanned character is an Operator and the stack is not empty, compare the precedence of the character with the element on top of the stack (topStack). If topStack has higher precedence over the scanned character Pop the stack else Push the scanned character to stack. Repeat this step as long as stack is not empty and topStack has precedence over the character.
  5. Repeat this step till all the characters are scanned.
  6. After all characters are scanned, we have to add any character that the stack may have to the Postfix string. If stack is not empty add topStack to Postfix string and Pop the stack. Repeat this step as long as stack is not empty.

The Code



Code: C

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>

#define N 64

#define LP 10
#define RP 20
#define OPERATOR 30
#define OPERAND 40

// Left parentheses precedence. Minimum of all
#define LPP 0

// Addition Subtraction precedence. Minimum among all operator precedence
#define AP 1
#define SP AP

// Multiplication divisor precedence.
#define MP 2
#define DP MP

// Remainder precedence.
#define REMP 2

#define NONE 9

static char infix[N+1],stack[N],postfix[N+1];
static int top;

void infixtopostfix(void);     /** POSTFIX CONVERSION FUNCTION **/
int gettype(char);             /** TYPE OF EXPRESSION GENERATOR **/
void push(char);               /** PUSH FUNCTION **/
char pop(void);                /** POP FUNCTION **/
int getprec(char);             /** PRECEDENCE CHECKER FUNCTION **/

void main()
{
    char ch;
    do
    {
        top=-1;
        printf("\nEnter an infix expression\n");
        fflush(stdin);
        gets(infix);
        infixtopostfix();
        printf("\ninfix = %s\npost fix =%s\n",infix,postfix);
        printf("\nDo you wish to continue\n");
        ch=getche();
    }while(ch=='Y' || ch=='y');
}

void infixtopostfix(void)
{
    int i,p,l,type,prec;
    char next;
    i=p=0;
    l=strlen(infix);
    while(i<l)
    {
        type=gettype(infix[i]);
        switch(type)
        {
        case LP:
            push(infix[i]);
            break;
        case RP:
            while((next=pop())!='(')
                postfix[p++]=next;
            break;
        case OPERAND:
            postfix[p++]=infix[i];
            break;
        case OPERATOR:
            prec=getprec(infix[i]);
            while(top>-1 && prec <= getprec(stack[top]))
                postfix[p++]=pop();
            push(infix[i]);
            break;
        }
        i++;
    }
    while(top>-1)
        postfix[p++]=pop();
    postfix[p]='\0';
}


int gettype(char sym)
{
    switch(sym)
    {
    case '(':
        return(LP);
    case ')':
        return(RP);
    case '+':
    case '-':
    case '*':
    case '/':
    case '%':
        return(OPERATOR);
    default :
        return(OPERAND);
    }
}

void push(char sym)
{
    if(top>N)
    {
        printf("\nStack is full\n");
        exit(0);
    }
    else
        stack[++top]=sym;
}

char pop(void)
{
    if(top<=-1)
    {
        printf("\nStack is empty\n");
        exit(0);
    }
    else
        return(stack[top--]);
}

int getprec(char sym)
{
    switch(sym)
    {
    case '(':
        return(LPP);
    case '+':
        return(AP);
    case '-':
        return(SP);
    case '*':
        return(MP);
    case '/':
        return(DP);
    case '%':
        return(REMP);
    default :
        return(NONE);
    }
}

Post Fix Expression Evaluation



Now after making the conversion what we have gained. As PostFix strings are parenthesis-free notation mathematical calculations and precedence is already defined within the string and so calculation is done very easily.

Postfix Expression evaluation Algo :
  1. Scan the Postfix string from left to right.
  2. Initialise an empty stack.
  3. If the scannned character is an operand, add it to the stack. If the scanned character is an operator, there will be atleast two operands in the stack.
  4. If the scanned character is an Operator, then we store the top most element of the stack(topStack) in a variable temp. Pop the stack. Now evaluate topStack(Operator)temp. Let the result of this operation be retVal. Pop the stack and Push retVal into the stack.
  5. Repeat this step till all the characters are scanned.
  6. After all characters are scanned, we will have only one element in the
    stack. Return topStack as result

The Code



Code: C

#include<stdio.h>
#include<conio.h>
#include<ctype.h>

#define N 100

void push(float,float **);
void pop(float **);

/**********************************************************/
/*****  Evaluation of generalised postfix expression  *****/
/*****       inputed as floating point numbers        *****/
/**********************************************************/

void main()
{
    int var,i,j;
    float stack[N];
    float *top;
    float num[N],res;
    char pfix[N],ch;
    do
    {
        j=0;
        top=stack;
        do
        {
            printf("\nEnter the total number of variables you will use\n");
            scanf("%d",&var);
            if(var>26)
                printf("\nMaximum 26 variables\n");
        }while(var>26);
        printf("\nAssign values to each of them\n");
        for(i=0;i<var;i++)
        {
            printf("\nAssign value to %c\t",('A'+i));
            scanf("%f",&num[i]);
        }
        printf("\nEnter a postfix expression using above variables\n");
        for(i=0;(pfix[i]=getche())!=13;i++)
        {
            if(islower(pfix[i])!=NULL)
            {
                pfix[i]=toupper(pfix[i]);
                printf("\b%c",pfix[i]);
            }
        }
        pfix[i]='\0';
        for(i=0;pfix[i]!='\0';i++)
        {
            if(isalpha(pfix[i])!=NULL)
                push(num[j++],&top);
            else
            {
                pop(&top);
                pop(&top);
                if(pfix[i]=='+')
                {
                    res=*top+*(top+1);
                    push(res,&top);
                }
                else if(pfix[i]=='-')
                {
                    res=*top-*(top+1);
                    push(res,&top);
                }
                else if(pfix[i]=='*')
                {
                    res=*top**(top+1);
                    push(res,&top);
                }
                else if(pfix[i]=='/')
                {
                    res=*top/(*(top+1));
                    push(res,&top);
                }
            }
        }
        printf("\nResult is %g\n",stack[0]);
        printf("\nDo you wish to continue[y/n]\n");
        ch=getche();
    }while(ch=='Y' || ch=='y');
    getch();
}

void push(float num,float **top)
{
    *(*top)=num;
    (*top)++;
}

void pop(float **top)
{
    (*top)--;
}

You can download both the codes in the attachment.

Peter_APIIT 20Apr2007 12:54

Re: InFix to PostFix and PostFix expression evaluation.
 
Nice Job

shabbir 20Apr2007 13:43

Re: InFix to PostFix and PostFix expression evaluation.
 
Quote:

Originally Posted by Peter_APIIT
Nice Job

Thanks.

bsnpr_24 24Apr2007 12:42

Re: InFix to PostFix and PostFix expression evaluation.
 
I would love to see the programs that changes infix to postfix and then evaluates it in C++ not in C. Could you give it to me or post it or soemthing i am anxious to see it, thanks!

shabbir 24Apr2007 13:51

Re: InFix to PostFix and PostFix expression evaluation.
 
Quote:

Originally Posted by bsnpr_24
I would love to see the programs that changes infix to postfix and then evaluates it in C++ not in C. Could you give it to me or post it or soemthing i am anxious to see it, thanks!

I guess the function can always be converted into class easily

bsnpr_24 24Apr2007 16:09

Re: InFix to PostFix and PostFix expression evaluation.
 
Here is what i got i need the main, i want two functions that you did the one that changes infix to post fix and one that evaluates, i am really stuck can you pleaseeee help me!

Here is my .h
Code:

#include <iostream>
using namespace std;

template <typename ElementType>
#ifndef STACK
#define STACK

class Stack
{
  private:
          class Node
          {
  public:
                ElementType info;
            Node *next;
          };
  typedef Node * Nodeptr;

    public:
                Stack();
                Stack(const Stack<ElementType> &);
                ~Stack();
                //Stack<ElementType>  operator=(const Stack<ElementType> &);
                bool empty() const;
                void Push(ElementType item);
                ElementType Pop();
                ElementType Top() const;
                void display(ostream & out) const;
               
        private:
                Node *miStack;
};
template <typename ElementType>
ostream & operator<< (ostream & out, const Stack <ElementType> & astack); //ya

#include "stack.template"
#endif

here is my template (I need something in the top function check it out)
Code:

#include<iostream>
using namespace std;

template <typename ElementType>
Stack<ElementType>::Stack()
{
        miStack=NULL;
}

template <typename ElementType>
Stack<ElementType>::Stack(const Stack<ElementType> & orig)
{
        nodePtr p=orig.miStack;
        nodePtr pnuevo, anterior;
        while (p!=NULL)
        {
                pnuevo=new node;
                pnuevo->info=p->info;
                if (orig.miStack==p)
                        miStack=pnuevo;
                else
                        anterior->next=pnuevo;
                anterior=pnuevo;
                p=p->next;
        }
        if (orig.miStack==NULL)
                miStack=NULL;
        else
                pnuevo->next=NULL;
}

template <typename ElementType>
Stack<ElementType>::~Stack()
{
        nodePtr p = miStack, anterior;
        while (p!=NULL)
        {
                anterior=p;
                p=p->next;
                delete anterior;
        }
}

/* template <typename ElementType>
Stack<ElementType> Stack<ElementType>::operator = (const Stack<ElementType> & orig)
{
        if (this != & orig)
        {
                nodePtr p=miStack, anterior;
                while (p!=NULL)
                {
                        anterior=p;
                        p=p->next;
                        delete anterior;
                }
        nodePtr q=orig.miStack;
        nodePtr pnuevo;
        while (q!=NULL)
        {
                pnuevo=new node;
                pnuevo->info=p->info;
                if (orig.miStack==p)
                        miStack=pnuevo;
                else
                        anterior->next=pnuevo;
                anterior=pnuevo;
                q=q->next;
        }
        if (orig.miStack==NULL)
                miStack=NULL;
        else
                miStack=NULL;
}
return *this;
}*/

template <typename ElementType>
bool Stack<ElementType>::empty() const
{
        return (miStack==NULL);
}

template <typename ElementType>
void Stack<ElementType>::Push(ElementType item)
{
        Nodeptr p = new Node;
        p->info=item;
        p->next=miStack;
        miStack=p;
}

template <typename ElementType>
ElementType Stack<ElementType>::Pop()
{
Nodeptr p = miStack, elementType item;
        if(miStack != null)
                item = p->info;
                miStack = miStack->next;
                delete p;
                return item;
        else
        {        cerr<<"Stack vacio.  No se hizo pop "<<endl;
                exit(1);
        }
}

template <typename ElementType>
ElementType Stack<ElementType>::Top()
{
        return( )//I dont know what to put here, help me also
}

template <typename ElementType>
void Stack<ElementType>::display (ostream & out)const
{
        nodePtr p;
        p=miStack;
        out<<"Los Elementos de la lista son:";
        while (p!=NULL)
        {
                out<<p->info<<" ";
                p=p->next;
        }
        cout<<endl;
}

template<typename ElementType>       
ostream & operator<<(ostream & out, const Stack<ElementType> & astack)

        astack.display(out);
        return out; 
}

here is the main(this is what i need help with):
Code:

#include "stack.h"

using namespace std;

void main()
{
       
//HELP!!

}

Please i need help ASAP

shabbir 24Apr2007 16:50

Re: InFix to PostFix and PostFix expression evaluation.
 
What you are trying to do is have some classes but don't know how to call them. How did you get the classes? If you got from somewhere I am sure they will have the sample implementation as well.

ranims 16Jun2007 16:59

Re: InFix to PostFix and PostFix expression evaluation.
 
if i am trying to use more than four variables in postfix expression, output is not coming to me.
for ex: ab+ cd+*
can u guide me
sorry if i am wrong

shabbir 17Jun2007 08:58

Re: InFix to PostFix and PostFix expression evaluation.
 
Copy the code into your compiler (I have used MS VC 6)
Run the Program and enter you will see the following text
Code:

Enter an infix expression
Enter the Infix expression as follows
Code:

(a+b)*(c+d)
You will see output as follows
Code:

infix = (a+b)*(c+d)
post fix =ab+cd+*

Do you wish to continue

I don't see any problem with the 4 variables as input. In fact I don't see any problem when I have input as
Code:

(a+b)*(c+d)*(e+f)
as well

ranims 18Jun2007 10:57

Re: InFix to PostFix and PostFix expression evaluation.
 
sorry i didn't mentioned clearly, While using postfix evaluation program, if i am typing postfix expression ab+cd+* , result is not coming properly, if ab+ result is ok.
can u verify once. where i am doing wrong

thanks in advance

shabbir 18Jun2007 11:01

Re: InFix to PostFix and PostFix expression evaluation.
 
I have used 4 variables as follows
A=2, B=3, C=4, D=5
with the expression as AB+CD+* and I got the following
Code:


Enter the total number of variables you will use
4

Assign values to each of them

Assign value to A      2

Assign value to B      3

Assign value to C      4

Assign value to D      5

Enter a postfix expression using above variables
AB+CD+*
Result is 45

. I don't see any problem. Can you please clarify a bit more.

abdo 9Sep2007 04:29

Re: InFix to PostFix and PostFix expression evaluation.
 
can you make this evaluates a postfix in java....
please cuz i saw it more easier for me to understand it.....
thank you

shabbir 9Sep2007 08:17

Re: InFix to PostFix and PostFix expression evaluation.
 
If you have understood it you should be able to convert it into any other language and then post it here for others to see.

keshaj 10Sep2007 06:48

Re: InFix to PostFix and PostFix expression evaluation.
 
i need a progrm in this....InFix to PostFix and PostFix expression evaluation.....program in c.... i really need ur help..

shabbir 10Sep2007 09:14

Re: InFix to PostFix and PostFix expression evaluation.
 
Quote:

Originally Posted by keshaj
i need a progrm in this....InFix to PostFix and PostFix expression evaluation.....program in c.... i really need ur help..

Did you read the article content. If not you should and you will get what you are looking for.

bloom_star7 15Sep2007 16:32

Re: InFix to PostFix and PostFix expression evaluation.
 
Shabbir,

In the 4th point you mentioed i think it should be operator not operand.

4.If the scanned character is an Operator[not Operand] and the stack is not empty, compare the precedence of the character with the element on top of the stack (topStack). If topStack has higher precedence over the scanned character Pop the stack else Push the scanned character to stack. Repeat this step as long as stack is not empty and topStack has precedence over the character.

May be i am wrong. Please let me konw if i missed something.

shabbir 15Sep2007 22:55

Re: InFix to PostFix and PostFix expression evaluation.
 
Yes it should be Operator as the precedence of the operand does not mean anything. Thanks for pointing it out and I have corrected that in the main article.

abdo 16Sep2007 23:13

Re: InFix to PostFix and PostFix expression evaluation.
 
i try to do this....i want to know if i use this program, is that correct?
Q:
How to write a java program taht evaluates a postfix expression by using concept of stack?
--------------------------------------------
Code:

public class PostfixEval
{

  public static void main(String[] args)
    {
     
      TextIO.putln("This program can evaluate postfix expressions such as\n");
      TextIO.putln("        2 2 +");
      TextIO.putln("or");
      TextIO.putln("        7.3 89.2 + 9 1.83 * 2 + /\n");
      TextIO.putln("The operators +, -, *, /, and ^ can be used.\n\n");
     
      while (true)
        {
            // Get and process one line of input from the user.
        TextIO.putln("\n\n\nEnter a postfix expression or press return to end:\n");
        TextIO.put("? ");
        skipSpaces();
        if (TextIO.peek() == '\n')
          {
              // If the line is empty (except for spaces), we are done.
            break;
        }
        readAndEvaluate(); // Process the input.
        TextIO.getln(); // Discard the end-of-line.
      }
     
      TextIO.putln("\n\nExiting program.");
     
  } // end main();
 
 
  static void skipSpaces()
    {
        // Skip past any spaces and tabs on the current input line.
        // When this routine returns, the next character is either
        // the end-of-line character, '\n', or is a non-blank.
      while (TextIO.peek() == ' ' || TextIO.peek() == '\t')
        {
        TextIO.getAnyChar();
      }
  }
 
 
  static void readAndEvaluate()
    {
        // Read one line of input and process it as a postfix expression.
        // If the input is not a legal postfix expression, then an error
        // message is displayed. Otherwise, the value of the expression
        // is displayed. It is assumed that the first character on
        // the input line is a non-blank. (This is checked in the
        // main() routine.)
     
      NumberStack stack; // For evaluating the expression.
     
      stack = new NumberStack(); // Make a new, empty stack.
     
      TextIO.putln();
     
      while (TextIO.peek() != '\n')
        {
     
          if ( Character.isDigit(TextIO.peek()) )
            {
                  // The next item in input is a number. Read it and
                  // save it on the stack.
              double num = TextIO.getDouble();
              stack.push(num);
              TextIO.putln("  Pushed constant " + num);
          }
          else
            {
                // Since the next item is not a number, the only thing
                // it can legally be is an operator. Get the operator
                // and perform the operation.
              char op; // The operator, which must be +, -, *, /, or ^.
              double x,y;    // The operands, from the stack, for the operation.
              double answer; // The result, to be pushed onto the stack.
              op = TextIO.getChar();
              if (op != '+' && op != '-' && op != '*' && op != '/' && op != '^')
                  {
                      // The character is not one of the acceptable operations.
                  TextIO.putln("\nIllegal operator found in input: " + op);
                  return;
              }
              if (stack.isEmpty())
                  {
                  TextIO.putln("  Stack is empty while trying to evaluate " + op);
                  TextIO.putln("\nNot enough numbers in expression!");
                  return;
              }
              y = stack.pop();
              if (stack.isEmpty())
                  {
                  TextIO.putln("  Stack is empty while trying to evaluate " + op);
                  TextIO.putln("\nNot enough numbers in expression!");
                  return;
              }
              x = stack.pop();
              switch (op)
                  {
                case '+': answer = x + y; break;
                case '-': answer = x - y; break;
                case '*': answer = x * y; break;
                case '/': answer = x / y; break;
                default:  answer = Math.pow(x,y); // (op must be '^'.)
              }
              stack.push(answer);
              TextIO.putln("  Evaluated " + op + " and pushed " + answer);
          }
         
          skipSpaces();
     
      } // end while
     
      // If we get to this point, the input has been read successfully.
      // If the expression was legal, then the value of the expression is
      // on the stack, and it is the only thing on the stack.
     
      if (stack.isEmpty())
        { // Impossible if the input is really non-empty.
          TextIO.putln("No expression provided.");
          return;
      }
     
      double value = stack.pop(); // Value of the expression.
      TextIO.putln("  Popped " + value + " at end of expression.");

      if (stack.isEmpty() == false) {
          TextIO.putln("  Stack is not empty.");
          TextIO.putln("\nNot enough operators for all the numbers!");
          return;
      }

      TextIO.putln("\nValue = " + value);
     
     
  } // end readAndEvaluate()


} // end class PostfixEval

-----------------------------------------------------------------------
Attached the question,i know for you is simple, but i really need your help...Thank you

shabbir 17Sep2007 09:26

Re: InFix to PostFix and PostFix expression evaluation.
 
Quote:

Originally Posted by abdo
Attached the question,i know for you is simple, but i really need your help...Thank you

You are posting Java question in a C-C++ article and I do not expect you would get any help relating to the same but anyway best of luck. You should have read Before you make a query ( Link in upper right corner ) for some better responses.

Now also about your query I think the code given in the article is simple enough for the Java programmer to convert it into Java but I am not a Java expert.

sarah24 27Sep2007 21:33

Re: InFix to PostFix and PostFix expression evaluation.
 
i need some help pls help me...
i have written a code for postfix expression evaluation but what if all the data and expressions to be evaluated are in text file..
pls tell me how can i modify my code ..
Code:

#include <iostream>
#include <stack>
#include <string>
using namespace std;

void main()
{
    int i, choice = 1;
    string postfixExp;
    char token;
    float value, value1, value2;
    stack<float> s; //Declare a stack of floats

 

    while (choice != 0)
    {
        cout << "1. Evaluate a postfix expression" << endl;
        cout << "0. Exit " << endl;
        cout << "Enter the number for the option: ";

        cin >> choice;
        switch(choice)
        {
            case 1: cout << "Evaluate a postfix expression\n";
                    cout << "Enter the expression: ";
                    cin >> postfixExp;
                    i = 0;
                    token = postfixExp[i];
                    while((i < postfixExp.size()) && (token != '='))
                    {
                        if(isdigit(token))
                        {
                            value = token - '0';
                            s.push(value);
                        }
                        else
                        {
                            value2 = s.top();
                            s.pop();
                            value1 = s.top();
                            s.pop();
                            switch(token)
                            {
                                case '+': value = value1 + value2;
                                          break;
                                case '-': value = value1 - value2;
                                          break;
                                case '*': value = value1*value2;
                                          break;
                                case '/': value = value1/value2;
                                          break;
                            }
                            s.push(value);
                        }
              i++;
              token = postfixExp[i];
                    }
                    value = s.top();
                    s.pop();
                    cout << postfixExp << " " << value << endl;       
                    break;

            case 0: cout << "Exiting the program\n";
                    break;

            default: cout << "Invalid option\n";
                    break;
        }
    cout << endl;
    }
}


sribalaji 6Oct2007 10:10

Re: InFix to PostFix and PostFix expression evaluation.
 
nice.very good

nirry4u2004 27Oct2007 08:01

Re: InFix to PostFix and PostFix expression evaluation.
 
urgent neeed of program infix to postfix convertion and evaluation using postfix in java using stacks........

ron2x 15Dec2007 13:23

Re: InFix to PostFix and PostFix expression evaluation.
 
gOod day sir....im new here and i was hOping if i could have an
answer to my question..im actually a cOmputer engineering student...
a sophomore but yet still having probs when it comes to programming...

in this program u made sir...is it possible not to use any brackets?
if yes how will the program code go now...
if not why should i or how am suppose to discuss why should every user
needs to put brackets...

was hoping not to put any brackets but still the output would be exact and accurate..

thanks sir...and Godspeed...:D

TC....

sunilkumarsheoran 4Nov2009 00:53

Re: InFix to PostFix and PostFix expression evaluation.
 
the problum with postfix solver is not been cured???

mohammad xxx 17Jul2010 20:51

Re: InFix to PostFix and PostFix expression evaluation.
 
Hello guys please help me, I want the code of the program "evaluation post fix expression" by the language of C + + Example:
5 3 9 * 6 1 - / + Vicu the final product is equal to 10
Please help I need to code tonight Thanks for all the

Khiine 19Aug2010 07:48

Re: InFix to PostFix and PostFix expression evaluation.
 
Hello there, i think there's something wrong with the code..
i entered the infix expression

A+(B*C-(D/E^F)*G)*H

the output of your code in postfix expression is this:

ABC*DE^F/G*-H*+ <- is this really correct?..


think there's a problem with the '^' and '/'..
im confused now..
which operator should be executed first?..


ABC*DEF^/G*-H*+ <- isn't it supposed to be like this?..

thanks..

shabbir 19Aug2010 09:13

Re: InFix to PostFix and PostFix expression evaluation.
 
^ is not supported in the above code.

applemanthra 14Sep2010 09:23

Re: InFix to PostFix and PostFix expression evaluation.
 
hi
could anyone can post the infix to prefix program using stacks in c++,and this should read the values from Resource File for ex:"values.txt"(the values shold be like
a 3
b 2
c 1
d 5
e 4
f 7
file and then after getting the postfix expression immediately it should display the total value of the infix/postfix expression...please can anyone help me...im not getting this program...its very very urgent...

debabrata 10Nov2010 12:54

Re: InFix to PostFix and PostFix expression evaluation.
 
Quote:

Originally Posted by shabbir (Post 11424)
Thanks.

The program is really nice & as it has been written in C language , it would be easier to understand to all

Pepper Mint 2Dec2012 07:01

Re: InFix to PostFix and PostFix expression evaluation.
 
Hi! I found this article really helpful and I'd like to extend my thanks to you. However, I was wondering if I would like to input numbers more than 9 (i.e 10, 23, 15), how could I possibly accomplish that? I would really be grateful on your response.

Jack Hard 20Dec2012 17:13

Re: InFix to PostFix and PostFix expression evaluation.
 
When I tried to use the expression in these four variables suffix, the output does not come from me.
For example, * AB + CD +
can guide
I'm sorry if I'm wrong


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