void main()
{
if(printf("semicolon")){}
if(printf(getch())){}
}

void main()
{
    while(!printf("\nHello World"))
}

Answer #1 and #2 are wrong. They dont work on Turbo C. If
anybody knows the write answer please email me the code.

#include<stdio.h>
void main()
{char a=';';
if(printf("%c",a)){}
}

#include<stdio.h>
void main()
{int a=59;
if(printf("%c",a)){}
}
\\the previous one was wrong

#include<stdio.h>
void main()
{
if(printf("%c",59)){}
}
\\actually, this is correct

Hey...thanx anshu....

ultimate

void main()
{
if(getche()){}
}
you can input ; at run time and itll be printed on output
screen.

main()
{
  if(printf("%c",59))
  {}
}

Anshus code is working,tanx

#include<stdio.h>
void main()
{
 while( printf("%c", 59), 0)
 {
 }
}

Does anyone know why its working??

("%c", 59)???

because 59 is the ascii value for a semicolon..

hence we are printing a character against a number..

This was juss as little information..
few may get confused.. :-)

#include<stdio.h>
void main()
{
 while( printf("%c", 59), 0)
 {
 }
}

int main()
{
 if(hellow world){}
}

main()
{
if(printf(";"))
{
}
}

#include<iostream>

using namespace std;

int main() {

    cout <<  char(59) << endl;

}

#include<stdio.h>
void main()
{
if(printf("semicolon")){}
if(getch()){}
}
this will print the word 'semicolon'

#include<stdio.h>
#include<conio.h>
void main()
{
if(printf("%c",59)){}
if(getch()){}
}
this is the perfect solution...