C Code, Can U Solve This? Another interesting problem

[cse-icons 0]

Hello,

Look at the code given below

void fun(void)[/br]{[br]/* Put your code here so that main does not print 20 */[/br]}[br][/br]int main()[br]{[/br] int i = 20;[br] fun();[/br] printf("i = %d\n", i);[br] return 0;[/br]}

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 0]

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 0]

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 0]

I haven't tested this code:

[br]void fun() {[br][/br]int aaaaaa;[/br]int *q= &aaaaaa;[br][/br]while ( *(q--)!=20 );   // we look for variable which value is 20 (it couldn't be 'i'!)[br][/br]*q = 10; // we change it[br][/br][br]}[br][/br]main() { ... }[/br]

Hm.... It is only suggestion.

[cse-icons 0]

good try tmonty...the solution wud work in a few cases, but just as i told earlier.. it cud failreason: some other memory location might contain "20"..chek out my next reply for a hint....

[nooc 0]

#include <stdio.h>[br][/br]void fun(void)[br]{[/br]#define fun() i=10[br]}[br][/br]int main()[br]{[/br]int i = 20;[/br]fun();[br]printf("i = %d\n", i);[/br]return 0;}

[cse-icons 0]

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

fun()[/br]{[br]#define printf( i , j ) printf( i , j+1 )[/br]}[br][/br]main ()[br]{[/br]int i =10;[br]fun();[/br]printf("i = %d",i);}

[switch 0]

very smooth.

[tmonty 0]

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

[jordanliuhao 0]

How about

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;}

[8bit 0]

void fun(void){  return 0;}

XD Who wants it to print 20 lets just end the program now!

[osknockout 0]

8bit, you can't do that (can have only 1 return 0;)Cse-icons you devil, who the hell woulddo that? Just playing around of course, Ididn't get it. I'm too practical

[8bit 0]

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

Beat that.

[ankitunlimited 0]

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

Beat that.

i dont know c, but c++. In c++, you cant access i anywhere else but main()

[fotomaniak 0]

void fun(){int i=0;while(true) i++;}overflow and program will end... or user will kill the program after waiting for a bit.--------------------------------void fun(){int *i=0;while(true) *(i++)=i;}same as above plus there is a chance that program corrupts some critical data and crashes