A puzzle in C:

Any solution of this prolbem:

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 */

}

void main()
{
      int i=5;

      change();

      i=10;

        printf("%d",i);
}

---

Sure.

Code:

---

#include<stdio.h>

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

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

---

Quote:

---

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.