C - proper syntax for pointer - c

Question

I call a function global var as follow:
char *Pointer;
I then pass it into function:
char *MyChar = DoSomething (&Pointer);
which is defined as:
char *DoSomething (char *Destination)
{
free (*Destination);
//re-allocate memory
Destination = malloc (some number);
//then do something...
//finally
return Destination;
}
it only works if I use (*Destination) instead of (Destination). can someone tell me if that is correct? I still do not understand why it does not take (Destination).

Answer 1

It is correct, Destination is already declared as a pointer, so you pass the address of Destination in DoSomething(&Destination), that is like a pointer to pointer, then you need to dereference Destination inside DoSomething() function, for which the indirection operator * works.
But the right way, is not to pass the address of the pointer, but the pointer instead like in
DoSomething(Destination);
now, since you want to malloc Destination inside the function, you should the do this
char * DoSomething( char **Destination )
{
// free( Destination ); why?
//re-allocate memory
*Destination = malloc( some number );
//then do something...
//finally
return *Destination;
}
this is a demonstration of how you can use pointers
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
char *copyString(const char *const source)
{
char *result;
int length;
length = strlen(source);
result = malloc(length + 1);
if (result == NULL)
return NULL;
strcpy(result, source);
printf("The address of result is : %p\n", result);
printf("The content of result is : %s\n", result);
printf("The first character of result is # %p\n", &result[0]);
return result;
}
int main()
{
char *string = copyString("This is an example");
printf("\n");
printf("The address of string is : %p\n", string);
printf("The content of string is : %s\n", string);
printf("The first character of string is # %p\n", &string[0]);
/* we used string for the previous demonstration, now we can free it */
free(string);
return 0;
}
if you execute the previous program, you will see that the pointers both point to the same memory, and the contents of the memory are the same, so calling free in main() will realease the memory.

Answer 2

Here is a correct approach
char *Pointer;
//,,, maybe allocating memory and assigning its address to Pointer
//... though it is not necessary because it is a global variable and
//... will be initialized by zero. So you may apply function free to the pointer.
char *MyChar = DoSomething( Pointer );
char * DoSomething( char *Destination )
{
free( Destination );
//re-allocate memory
Destination = malloc( some number );
//then do something...
//finally
return Destination;
}
As for your code then
Type of the argument does not correspond to type of the parameter in function call
char *MyChar = DoSomething (&Pointer);
the type of the parameter is char * ( char *Destination ) while the type of argument is
char ** ( &Pointer )
As Destination is a pointer then instead of
free (*Destination);
you have to write
free( Destination );

Answer 3

It's because you are passing in an address of the pointer char *Pointer with the line
char *MyChar = DoSomething (&Pointer);
Since you are passing in the address of the pointer in your function DoSomething it sees the functional scope variable Destination as a pointer to an address that is the address of the pointer Pointer.
So rather than passing in the address of Pointer with
char *MyChar = DoSomething(&Pointer);
you need to pass in the pointer itself like so:
char *MyChar = DoSomething(Pointer);
which will allow you to use
free(Destination);
Notice the lack of & indicating the address of Pointer.