Let I be an algebraic expression written in infix notation. I may contain parentheses, operands, and operators. For simplicity of the algorithm we will use only +, –, *, /, % operators. The precedence of these operators can be given as follows:
Higher priority *, /, %
Lower priority +, –
No doubt, the order of evaluation of these operators can be changed by making use of parentheses. For example, if we have an expression A + B * C, then first B * C will be done and the result will be added to A. But the same expression if written as, (A + B) * C, will evaluate A + B first and then the result will be multiplied with C
The algorithm uses a stack to temporarily hold operators. The postfix expression is obtained from left-to-right using the operands from the infix expression and the operators which are removed from the stack. The first step in this algorithm is to push a left parenthesis on the stack and to add a corresponding right parenthesis at the end of the infix expression. The algorithm is repeated until the stack is empty
Algorithm to convert an infix notation to postfix notation
Program to convert an infix expression into its equivalent postfix notation using C
#include <stdio.h>
#include <conio.h>
#include <ctype.h>
#include <string.h>
#define MAX 100
char st[MAX];
int top=–1;
void push(char st[], char);
char pop(char st[]);
void InfixtoPostfix(char source[], char target[]);
int getPriority(char);
int main()
{
char infix[100], postfix[100];
clrscr();
printf("\n Enter any infix expression : ");
gets(infix);
strcpy(postfix, "");
InfixtoPostfix(infix, postfix);
printf("\n The corresponding postfix expression is : ");
puts(postfix);
getch();
return 0;
}
void InfixtoPostfix(char source[], char target[])
{
int i=0, j=0;
char temp;
strcpy(target, "");
while(source[i]!='\0')
{
if(source[i]=='(')
{
push(st, source[i]);
i++;
}
else if(source[i] == ')')
{
while((top!=–1) && (st[top]!='('))
{
target[j] = pop(st);
j++;
}
if(top==–1)
{
printf("\n INCORRECT EXPRESSION");
exit(1);
}
temp = pop(st);//remove left parenthesis
i++;
}
else if(isdigit(source[i]) || isalpha(source[i]))
{
target[j] = source[i];
j++;
i++;
}
else if (source[i] == '+' || source[i] == '–' || source[i] == '*' ||
source[i] == '/' || source[i] == '%')
{
while( (top!=–1) && (st[top]!= '(') && (getPriority(st[top])
> getPriority(source[i])))
{
target[j] = pop(st);
j++;
}
push(st, source[i]);
i++;
}
else
{
printf("\n INCORRECT ELEMENT IN EXPRESSION");
exit(1);
}
}
while((top!=–1) && (st[top]!='('))
{
target[j] = pop(st);
j++;
}
target[j]='\0';
}
int getPriority(char op)
{
if(op=='/' || op == '*' || op=='%')
return 1;
else if(op=='+' || op=='–')
return 0;
}
void push(char st[], char val)
{
if(top==MAX–1)
printf("\n STACK OVERFLOW");
else
{
top++;
st[top]=val;
}
}
char pop(char st[])
{
char val=' ';
if(top==–1)
printf("\n STACK UNDERFLOW");
else
{
val=st[top];
top––;
}
return val;
}
Output
Enter any infix expression : A+B–CD
The corresponding postfix expression is : AB+CD–