use of swap with pointers gave error.

ofbodur · 05-13-2013, 10:23 AM

Dear visitors,

I typed the code below to my text editor and executed by using gcc compiler:

Code:

#include<stdio.h>
main()
{     int a,b,c;
      scanf(“%d*“,&a);
      scanf(“*%d*“,&b);
      printf(“%d %d \n“,a,b);
       swap(&a, &b);
       printf(“%d %d \n”,a,b);
}
/*This is the C subroutine that does the swap using pointers.*/
void swap(int *px, int *py)  
/* interchange *px and *py */
{  int temp;
temp = *px;
*px = *py;
*py = temp;   
}

but, it gave me the error below:

new.c:11:6: warning: conflicting types for ‘swap’ [enabled by default]
new.c:7:8: note: previous implicit declaration of ‘swap’ was here

what is the problem here?
Any further help is muchly appreciated..

johnsfine · 05-13-2013, 11:26 AM

You need to declare the function before calling it.

Code:

#include<stdio.h>
void swap(int *px, int *py);
main()
{     int a,b,c;
      scanf(“%d*“,&a);
      scanf(“*%d*“,&b);
      printf(“%d %d \n“,a,b);
       swap(&a, &b);
       printf(“%d %d \n”,a,b);
}
/*This is the C subroutine that does the swap using pointers.*/
void swap(int *px, int *py)  
/* interchange *px and *py */
{  int temp;
temp = *px;
*px = *py;
*py = temp;   
}

ofbodur · 05-13-2013, 11:39 AM

is it also OK to declare void swap(); in the main function?

NevemTeve · 05-13-2013, 10:56 PM

possible, but not recommended:

Code:

int main (void)
{
    void swap (int *, int *);
...

sundialsvcs · 05-16-2013, 10:11 AM

Agreed. The "C" language allows such declarations to be made in a per-function limited scope, but that's not normally what is done. The normal practice in that language is to issue declarations in advance, so that they have a per-file-level scope. (Case in point is stdio.h itself, which, if you care to open the file and look at it, is chock-full of just such declarations, concerning the standard-I/O library.) The preferred practice, "when in Rome," is "to do what the Romans do."