Postfix to Infix Conversion
Algorithm of Postfix to Infix
Expression = abc-+de-fg-h+/*
1.While there are input symbol left
2. Read the next symbol from input.
3. If the symbol is an operand
Push it onto the stack.
4. Otherwise,
the symbol is an operator.
5. If there are fewer than 2 values on the stack
Show Error /* input not sufficient values in the expression */
6. Else
Pop the top 2 values from the stack.
Put the operator, with the values as arguments and form a string.
Encapsulate the resulted string with parenthesis.
Push the resulted string back to stack.
7. If there is only one value in the stack
That value in the stack is the desired infix string.
8. If there are more values in the stack
Show Error /* The user input has too many values */
Postfix to Infix conversion
Example
abc-+de-fg-h+/*
| Expression | Stack | ||||
|---|---|---|---|---|---|
| abc-+de-fg-h+/* | NuLL | ||||
| bc-+de-fg-h+/* | "a" | ||||
| c-+de-fg-h+/* |
| ||||
| -+de-fg-h+/* |
| ||||
| +de-fg-h+/* |
| ||||
| de-fg-h+/* |
| ||||
| e-fg-h+/* |
| ||||
| -fg-h+/* |
| ||||
| fg-h+/* |
| ||||
| g-h+/* |
| ||||
| -h+/* |
| ||||
| h+/* |
| ||||
| +/* |
| ||||
| /* |
| ||||
| * |
| ||||
| Null |
|
Write a Program to Implement Postfix to Infix in c
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
int top = 10;
struct node
{
char ch;
struct node *next;
struct node *prev;
} *stack[11];
typedef struct node node;
void push(node *str)
{
if (top <= 0)
printf("Stack is Full ");
else
{
stack[top] = str;
top--;
}
}
node *pop()
{
node *exp;
if (top >= 10)
printf("Stack is Empty ");
else
exp = stack[++top];
return exp;
}
void convert(char exp[])
{
node *op1, *op2;
node *temp;
int i;
for (i=0;exp[i]!='\0';i++)
if (exp[i] >= 'a'&& exp[i] <= 'z'|| exp[i] >= 'A' && exp[i] <= 'Z')
{
temp = (node*)malloc(sizeof(node));
temp->ch = exp[i];
temp->next = NULL;
temp->prev = NULL;
push(temp);
}
else if (exp[i] == '+' || exp[i] == '-' || exp[i] == '*' || exp[i] == '/' ||
exp[i] == '^')
{
op1 = pop();
op2 = pop();
temp = (node*)malloc(sizeof(node));
temp->ch = exp[i];
temp->next = op1;
temp->prev = op2;
push(temp);
}
}
void display(node *temp)
{
if (temp != NULL)
{
display(temp->prev);
printf("%c", temp->ch);
display(temp->next);
}
}
void main()
{
char exp[50];
clrscr();
printf("Enter the postfix expression :");
scanf("%s", exp);
convert(exp);
printf("\nThe Equivalant Infix expression is:");
display(pop());
printf("\n\n");
getch();
}
Output:
Enter the postfix expression: abc-+
The Equivalent infix expression is: a+b-c
