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Conversion of postfix expression to infix

How do we convert a postfix expression to infix expression?

A postfix expression or a "Reverse polish notation" is useful for programming. Because our programs can easily evaluate a postfix expression.

But for humans, postfix expressions are difficult to understand.

So is it possible to convert a given postfix expression to infix expression? Why not? It is possible with the help of stack data structure. 

Remember that when converting an expression from infix to postfix, we used operator stack. But in this case we need an operand stack.

What we need to do is - we extract values from expression. If there is an operand, we push it to stack. If there is an operator encountered, we pop two most recent values from stack, apply operator to them, enclose them in paranthesis and push the expression back to stack.

This procedure is continued until the entire expression is scanned. In the next step, the content of stack is popped out - which will be our infix expression.

So let me bullet these statements
  1. scan a character from postfix expression.
  2. if the character is an operand, push it to stack
  3. if the character scanned is an operator 
    1. pop expr1
    2. pop expr2
    3. form expr3 as (expr1 operator expr2)
  4. push expr3 to stack
  5. Repeat the steps 1 to 3, until all characters are scanned 
  6. pop content of stack as infix expression 
Let us look at an example postfix expression  53*4+

characterstackComments
5 5
35 3
*5  3 is popped
      5 is popped
     (5*3) is pushed
(5*3) 
4(5*3) 4
+ (5*3) - 4 is popped    
*   (5*3) is popped
  ((5*3)+4) is pushed
((5*3)+4)

Next the stack has ( (5*3)+4) which is popped. And that is our infix expression.

Let us write our C function for this now.


void convert(char *postfix,char *infix)
{
     char ch;
     char oper1,oper2,opernew;
     char *expr1=(char*)malloc(30);
     char *expr2=(char*)malloc(30);
     char *expr3 = (char*)malloc(60);
     struct node *top = NULL;
     while(ch =*postfix++)
     {
 if(isdigit(ch))
        {
            char  str[] = {ch,0};
     top = push(str,top);
        }
        else if(is_operator(ch))
 {
    strcpy(expr1, pop(&top));
    if(expr1==NULL)
   {
                   printf("Error");
     break;
                 }
    strcpy(expr2 ,pop(&top));
    if(expr2==NULL)
   {
                   printf("Error");
     break;
                 }
           strcpy(expr3, concat(expr2,expr1,ch));    
           top = push(expr3,top);
          }
 } 
        strcpy(infix,pop(&top));
}


And here is the driver program. As you can see from the code, I have used linked list implementation of stack of strings here.


#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
struct node
{
    char str[30];
    struct node *next;
};
struct node *createnode(char *str)
{
    struct node *newnode = (struct node*) malloc(sizeof(struct node));
    strcpy(newnode->str,str);
    newnode->next = NULL;
}

struct node * push(char *str,struct node *top)
{
   struct node *newnode = createnode(str);
   newnode->next = top;
   top = newnode;
   return top;
}

char * pop(struct node ** top)
{
   static char str[30];
   if(top==NULL)
 return NULL;
   struct node * temp = *top;
   *top =(*top)->next;
   strcpy(str,temp->str);
   
   free(temp);
   return str;
}

int is_operator(char ch)
{
    switch(ch)
    {
 case '+':
 case '-':
 case '/':
 case '*':
 case '^':
  return 1;
        default : return 0;
    }
}
 
char *concat(char *s1,char *s2,char ch)
{
   char *result = (char*)malloc(60);
   int i=1;
   result[0] = '(';
   while(*s1)
     result[i++]=*s1++;
   result[i++]=ch;
   while(*s2)
     result[i++]=*s2++;
   result[i++]=')';//enclose in parantheses
   result[i]=0;
   return result;
}
void convert(char *postfix,char *infix)
{
     char ch;
     char oper1,oper2,opernew;
     char *expr1=(char*)malloc(30);
     char *expr2=(char*)malloc(30);
     char *expr3 = (char*)malloc(60);
     struct node *top = NULL;
     while(ch =*postfix++)
     {
 if(isdigit(ch))
        {
            char  str[] = {ch,0};
     top = push(str,top);
        }
        else if(is_operator(ch))
 {
    strcpy(expr1, pop(&top));
    if(expr1==NULL)
   {
                   printf("Error");
     break;
                 }
    strcpy(expr2 ,pop(&top));
    if(expr2==NULL)
   {
                   printf("Error");
     break;
                 }
           strcpy(expr3, concat(expr2,expr1,ch));    
           top = push(expr3,top);
          }
 }
        //strcpy(expr3 = pop(&top);
        strcpy(infix,pop(&top));
}
    
   
int main()
{
   char postfix[30];char infix[30];
   double ans;
   printf("Enter postfix expression :");
   scanf("%s",postfix); 
   convert(postfix,infix);
   printf("The   expression in infix is %s",infix);
   return 0;
}

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