Understanding pass pointer to pointer by reference in c

[surfersss]

void func(char *ptr)        //Passed by      reference
{
  *ptr = 'B';
}

int main()
{
  char *ptr;
  ptr = (char *) malloc(sizeof(char) * 1);
  *ptr = 'A';
  printf("%c\n", *ptr);

  func(ptr);
  printf("%c\n", *ptr);
}

I was confuse I know that ptr=(char*)malloc (sizeof(char*1) here means allocate 1 byte and return a pointer to the allocation then assign to ptr, so ptr is a pointer.

But when it calls func(ptr) why it not use &? Although what I want is to change the character inside ptr points to? Why not using void func(char** ptr) and send func(&ptr) here? Is it possible?

I mean what made this pass by reference and not by value?
Why this not pass copy of the pointer?
Is this mean in func(ptr), i pass pointer which is memory address(malloc) and was define as char*ptr in func?? So thats why it work?

[Peter87]

The pointer is passed by value. The function doesn't modify the pointer, it modifies the character that the pointer points to.