Thread: A puzzle in C:

Sure.

Code:

#include<stdio.h>

void change()
{
#define printf i=5,printf
}

void main()
{
	int i=5;
	
	change();
	
	i=10;
	
	printf("%d",i);
}

Originally Posted by pianorain

Sure.

Code:

#include<stdio.h>

void change()
{
#define printf i=5,printf
}

void main()
{
	int i=5;
	
	change();
	
	i=10;
	
	printf("%d",i);
}

Wow, would of never thought of that one. I guess that's one of those classic obfuscation tricks.

Only way I knew how to do it. I don't usually use #define's in obfuscation. I think it's kind of cheap.

Code:

void change( void )
{
    printf("5");
    fflush( stdout );
    exit( 0 );
}

:P


Quzah.

Your solution won't work, quzah, because the statement requires that the printf statment in main() will result in printing "5." Nice try though.

Here's my thought, along the same lines as pianorain's:

Code:

#include <stdio.h>

void change() {
/* Write something in this function so that the output of printf
   in mainfunction should give 5 . Do not change the main function */
  #define printf(a, x) putchar('5')
}

int main() {
  int i = 5;
  change();
  i = 10;
  
  printf("%d",i);
  
  return 0;
}

I know. I was faking it. Hence the ':P'.


Quzah.

> Any solution of this prolbem:
Yeah, find a better book or teacher.