Question

In this program I have swapped the first 2 names

#include<stdio.h>
void swap(char **,char **);
main()
{
 char *name[4]={"amol", "robin", "shanu" };
 swap(&name[0],&name[2]);
 printf("%s %s",name[0],name[2]);
}
void swap(char **x,char **y)
 {
 char *temp;
 temp=*x;
 *x=*y;
 *y=temp;
 }

This programs runs perfectly but when I use the function swap(char *,char *) it does not swap the address why? why I have to use pointer to pointer?

Answer 1

I assume you understand that to swap integers you would have function like swap(int *, int *)

Similarly, When you want to swap strings which is char *. You would need function like swap(char **, char **).

In such cases, you take their pointers and swap their content (otherwise values will not be swapped once function returns). For integer content, pointer is int * and in case of strings content is char * pointer to it is char **.