Using pointers to communicate between functions - re char *

We all know that we can change the values of two integers x and y using the follwing code:

Code:

---

void interchange(int *u, int *v) {
  int temp;
  temp = *u;
  *u = *v;
  *v = temp;
}
int main() {
  int x=5, y= 10;
  interchange( &x, &y);
}

---

That is, you pass the address of x (&x) to a function that has a parameter *x.
What if x is itself a pointer?

Code:

---

void func ( char **array1, char *array ) {
  array1 = array;
}
int main ( ) {
  char *msg;
  char *msg1;
  msg = malloc(20);
  sprintf(msg, "hello world");
  func( &msg1, msg );
}

---

What I really want to achieve is msg1 = msg. But gcc isssues two warnings:
1. " assignment from incompatible pointer type" for 'array1 = array'
2. "passing arg 1 of `func' from incompatible pointer type" for 'func( &msg1, msg )'
And if I print msg1 at the end, I get a segmentation fault.

[Edit]Ooops. I got it. I should have said *array1 = array;
Now there is no seg fault. But warning #2 remains. Why?

No warning appeared to me.Im using gcc 3.4.3 on RHEL4,even I turned -Wall option.

Code:

---

#include<stdio.h>

void func ( char **array1, char *array ) {
  *array1 = array;
}
int main ( ) {
  char *msg;
  char *msg1;
 
  msg = malloc(20);
  msg1 = malloc(20);
 
  sprintf(msg, "hello world");
  func( &msg1, msg );
  printf("%s",msg1);
 
  return 0;
}
/*myoutput
hello world
*/

---

ssharish2005

I have'nt allocated memory for msg1,then also it gives same output.As Im not accessing further elements from msg1 it works.