Thread: 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?