@Anika Jain
#include#define FUN(a,b,c)c t; t=a;a=b;b=t;
int main(){
int a=2,b=3;
int *x,*y;
x=&a,y=&b;
printf("%d %d \n",a,b);
printf("%u %u \n",x,y);
FUN(x,y,int *);
printf("%d %d \n",a,b);
printf("%u %u \n",x,y);
return 0;}
Output:
1234
2 3
32128519
You're printing a & b.
Value of a and b are not changed.
On Wed, Jun 29, 2011 at 3:51 PM, Naveen Kumar
wrote:
> Its swapping correctly
> Now x is pointing to &b and y is pointing to &a.
>
> On Wed, Jun 29, 2011 at 3:45 PM, Anika Jain wrote:
>>
>> #define FUN(a,b,c) c t; t=a;a=b;b=t;
>>
>>
>>
Its swapping correctly
Now x is pointing to &b and y is pointing to &a.
On Wed, Jun 29, 2011 at 3:45 PM, Anika Jain wrote:
>
> #define FUN(a,b,c) c t; t=a;a=b;b=t;
>
>
> int main()
> {
> int a=2,b=3;
> int *x,*y;
> x=&a,y=&b;
> FUN(x,y,int *);
> printf("%d %d \n",a,b);
>
#define FUN(a,b,c)c t; t=a;a=b;b=t;
int main()
{
int a=2,b=3;
int *x,*y;
x=&a,y=&b;
FUN(x,y,int *);
printf("%d %d \n",a,b);
return 0;
}
why doesnt this macro swap pointers wen it can work for variables??
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