C Code, Can U Solve This? - Another interesting problem

cse-icons

Nov 17 2004, 07:07 AM

Hello,

Look at the code given below

CODE

void fun(void)
{
/* Put your code here so that main does not print 20 */
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

1. You are allowed to put your code ONLY in fun.
2. You are not allowed to change even one character in main.
3. You should not use functions like exit(),abort() in the function fun.

How do I manipulate and change the value of i?
Have fun...

no9t9

Nov 17 2004, 12:55 PM

This smells like a homework assignment... So.. I won't tell you how to do it

Here's a POINTER though... variables can be manipulated using their address in memory...

cse-icons

Nov 18 2004, 04:51 AM

actually, if u use pointers, u cannot be sure that it will work on all systems.
coz memory allocation is compiler and architecture specific....

btw, this is not an assignment and also I got the answer without using pointers....
just give it a shot....

tmonty

Nov 18 2004, 04:15 PM

I haven't tested this code:

CODE

void fun() {

int aaaaaa;
int *q= &aaaaaa;

while ( *(q--)!=20 ); // we look for variable which value is 20 (it couldn't be 'i'!)

*q = 10; // we change it

}

main() { ... }

Hm.... It is only suggestion.

cse-icons

Nov 19 2004, 07:35 AM

good try tmonty...
the solution wud work in a few cases, but just as i told earlier.. it cud fail
reason: some other memory location might contain "20"..

chek out my next reply for a hint....

nooc

Nov 19 2004, 10:02 PM

CODE

#include <stdio.h>

void fun(void)
{
#define fun() i=10
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

cse-icons

Nov 22 2004, 08:34 AM

yeah, that cud be a solution.... i have not tried out...
My solution is also on similar lines..... here it is:

CODE

fun()
{
#define printf( i , j ) printf( i , j+1 )
}

main ()
{
int i =10;
fun();
printf("i = %d",i);
}

switch

Nov 26 2004, 02:38 AM

very smooth.

tmonty

Nov 26 2004, 04:47 PM

Yeah. ;p
With #define U can do everything. ;p

jordanliuhao

Feb 1 2005, 02:02 AM

How about

CODE

void fun(void)
{
/* Put your code here so that main does not print 20 */
}

void printf(....)
{
/* Above is my code, Just for fun :D */

}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

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tricky77puzzle

Feb 27 2008, 12:41 AM

printf is a standard function in C. You can't modify it.

Try doing the same thing in c++.

Dark_Prisoner

Feb 26 2008, 07:15 PM

QUOTE(nooc @ Nov 19 2004, 11:02 PM)

CODE

#include <stdio.h>

void fun(void)
{
#define fun() i=10
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

I think that this won't work :
(i guess) we can't put preprocessor directives on a function
If we can , it won't work also , because the compiler or the processor had completed the line :

CODE

fun();

so he won't replace it .

if the preprocessor works in functions i guess this will be the answer

CODE

void fun(void)
{
#define printf("i = %d\n", i) i=0
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

What do you think ???

klusk

Feb 25 2008, 02:26 PM

QUOTE(8bit @ Feb 4 2005, 07:07 AM)

CODE

void fun(void)
{
int * ptr =&i;
*ptr=50;
}

Beat that.

This code wont work as the variable i is undeclared for fun()....it would have worked if "i" had been a global variable

There is another way if you wish to use pointers...

CODE

void fun(int *ptr)
{
*ptr=50;
}

int main()
{
int i = 20;
fun(&i);
printf("i = %d\n", i);
return 0;
}

the output will be 50

Csshih

Dec 13 2007, 06:12 PM

Oh yea?
void fun(void)
{
randomlineofetext(3249830325)
}
the program won't print anything at all xD

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