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Title Convert infix expression to postfix expression
Author Mahesh.A.R.
Author Email maheshar2002 [at] yahoo.com
Description This program converts an infix expression to its
corresponding postfix expression

Category C » Beginners / Lab Assignments
Hits 364001
Code Select and Copy the Code
Code : #include<stdio.h> #include<conio.h> int top=0; char s[50]; int opr(char); void put(char); void main() { char a[50],b[50],ele,x; int y,z,r,j,i=1,m=0; clrscr(); printf(" Enter the infix expression : "); while(i<50) { scanf("%c",&a[i]); if(a[i]==10) break; i++; } y=i; a[0]='('; a[y]=')'; i=0; put(a[0]); while(i<y) { i++; ele=a[i]; x=s[top--]; if(ele>='a'&&ele<='z') { b[m++]=ele; put(x); } else { if(ele==')') while(x!='(') { b[m++]=x; x=s[top--]; } else { if(ele=='(') { put(x); put(ele); } else { z=opr(x); r=opr(ele); if(z>=r) { while(z>=r) { b[m++]=x; x=s[top--]; z=opr(x); } put(x); put(ele); } else { put(x); put(ele); }}}}} printf(" The postfix expression is : "); for(i=0;i<=(m-1);i++) printf("%c",b[i]); getch(); } void put(char ele) { top++; s[top]=ele; } int opr(char c) { int z; switch(c) { case '+':z=3; break; case '-':z=3; break; case '*':z=5; break; case '/':z=5; break; case '^':z=6; break; case '(':z=1; break; } return(z); }

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