Q:

# Infix To Postfix Conversion Using Stack [with C program]

## Infix Expression

It follows the scheme of <operand><operator><operand> i.e. an <operator> is preceded and succeeded by an <operand>. Such an expression is termed infix expression. E.g., A+B

## Postfix Expression

It follows the scheme of <operand><operand><operator> i.e. an <operator> is succeeded by both the <operand>. E.g., AB+

### Algorithm to convert Infix To Postfix

Let, X is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression Y.

1. Push “(“onto Stack, and add “)” to the end of X.
2. Scan X from left to right and repeat Step 3 to 6 for each element of X until the Stack is empty.
3. If an operand is encountered, add it to Y.
4. If a left parenthesis is encountered, push it onto Stack.
5. If an operator is encountered ,then:
1. Repeatedly pop from Stack and add to Y each operator (on the top of Stack) which has the same precedence as or higher precedence than operator.
[End of If]
6. If a right parenthesis is encountered ,then:
1. Repeatedly pop from Stack and add to Y each operator (on the top of Stack) until a left parenthesis is encountered.
2. Remove the left Parenthesis.
[End of If]
[End of If]
7. END.

Let’s take an example to better understand the algorithm

Infix Expression: A+ (B*C-(D/E^F)*G)*H, where ^ is an exponential operator.

Resultant Postfix Expression: ABC*DEF^/G*-H*+

### Advantage of Postfix Expression over Infix Expression

An infix expression is difficult for the machine to know and keep track of precedence of operators. On the other hand, a postfix expression itself determines the precedence of operators (as the placement of operators in a postfix expression depends upon its precedence).Therefore, for the machine it is easier to carry out a postfix expression than an infix expression.

### C program to convert Infix to Postfix Expression

``````/* This program converts infix expression to postfix expression.
* This program assume that there are Five operators: (*, /, +, -,^)
in infix expression and operands can be of single-digit only.
* This program will not work for fractional numbers.
* Further this program does not check whether infix expression is
valid or not in terms of number of operators and operands.*/

#include<stdio.h>
#include<stdlib.h>      /* for exit() */
#include<ctype.h>     /* for isdigit(char ) */
#include<string.h>

#define SIZE 100

/* declared here as global variable because stack[]
* is used by more than one fucntions */
char stack[SIZE];
int top = -1;

/* define push operation */

void push(char item)
{
if(top >= SIZE-1)
{
printf("\nStack Overflow.");
}
else
{
top = top+1;
stack[top] = item;
}
}

/* define pop operation */
char pop()
{
char item ;

if(top <0)
{
printf("stack under flow: invalid infix expression");
getchar();
/* underflow may occur for invalid expression */
/* where ( and ) are not matched */
exit(1);
}
else
{
item = stack[top];
top = top-1;
return(item);
}
}

/* define function that is used to determine whether any symbol is operator or not
(that is symbol is operand)
* this fucntion returns 1 if symbol is opreator else return 0 */

int is_operator(char symbol)
{
if(symbol == '^' || symbol == '*' || symbol == '/' || symbol == '+' || symbol =='-')
{
return 1;
}
else
{
return 0;
}
}

/* define fucntion that is used to assign precendence to operator.
* Here ^ denotes exponent operator.
* In this fucntion we assume that higher integer value
* means higher precendence */

int precedence(char symbol)
{
if(symbol == '^')/* exponent operator, highest precedence*/
{
return(3);
}
else if(symbol == '*' || symbol == '/')
{
return(2);
}
else if(symbol == '+' || symbol == '-')          /* lowest precedence */
{
return(1);
}
else
{
return(0);
}
}

void InfixToPostfix(char infix_exp[], char postfix_exp[])
{
int i, j;
char item;
char x;

push('(');                               /* push '(' onto stack */
strcat(infix_exp,")");                  /* add ')' to infix expression */

i=0;
j=0;
item=infix_exp[i];         /* initialize before loop*/

while(item != '\0')        /* run loop till end of infix expression */
{
if(item == '(')
{
push(item);
}
else if( isdigit(item) || isalpha(item))
{
postfix_exp[j] = item;              /* add operand symbol to postfix expr */
j++;
}
else if(is_operator(item) == 1)        /* means symbol is operator */
{
x=pop();
while(is_operator(x) == 1 && precedence(x)>= precedence(item))
{
postfix_exp[j] = x;                  /* so pop all higher precendence operator and */
j++;
x = pop();                       /* add them to postfix expresion */
}
push(x);
/* because just above while loop will terminate we have
oppped one extra item
for which condition fails and loop terminates, so that one*/

push(item);                 /* push current oprerator symbol onto stack */
}
else if(item == ')')         /* if current symbol is ')' then */
{
x = pop();                   /* pop and keep popping until */
while(x != '(')                /* '(' encounterd */
{
postfix_exp[j] = x;
j++;
x = pop();
}
}
else
{ /* if current symbol is neither operand not '(' nor ')' and nor
operator */
printf("\nInvalid infix Expression.\n");        /* the it is illegeal  symbol */
getchar();
exit(1);
}
i++;

item = infix_exp[i]; /* go to next symbol of infix expression */
} /* while loop ends here */
if(top>0)
{
printf("\nInvalid infix Expression.\n");        /* the it is illegeal  symbol */
getchar();
exit(1);
}
if(top>0)
{
printf("\nInvalid infix Expression.\n");        /* the it is illegeal  symbol */
getchar();
exit(1);
}

postfix_exp[j] = '\0'; /* add sentinel else puts() fucntion */
/* will print entire postfix[] array upto SIZE */

}

/* main function begins */
int main()
{
char infix[SIZE], postfix[SIZE];         /* declare infix string and postfix string */

/* why we asked the user to enter infix expression
* in parentheses ( )
* What changes are required in porgram to
* get rid of this restriction since it is not
* in algorithm
* */
printf("ASSUMPTION: The infix expression contains single letter variables and single digit constants only.\n");
printf("\nEnter Infix expression : ");
gets(infix);

InfixToPostfix(infix,postfix);                   /* call to convert */
printf("Postfix Expression: ");
puts(postfix);                     /* print postfix expression */

return 0;
}``````

Output

```First Run:
Enter Infix expression : A+(B*C-(D/E^F)*G)*H
Postfix Expression: ABC*DEF^/G*-H*+

Second Run:
Enter Infix expression : (3^2*5)/(3*2-3)+5
Postfix Expression: 32^5*32*3-/5+
```