Pointer initialization: address or value

Question

I have a question regarding pointer initialization in C.

I understand that *ptr will give the value of that pointer is pointing to.

ptr will give you the address.

Now I got following syntax:

int *ptr = (int *) malloc(sizeof(*ptr));

Why is *ptr being initialized with an address of the Heap and not a value? malloc() returns an address right?

Shouldn't it be:

int *ptr;
ptr = malloc(...);

Answer 1

With *ptr, * is acting as the dereferencing operator.

With int *ptr, * is acting as part of the type declaration for ptr.

So the two things are entirely different, even though * is used. (Multiplication and comment blocks are further uses of * in C).

Answer 2

In that line, int * is the type.

int *ptr = (int *) malloc(sizeof(*ptr));

Is just this compressed into one line:

int *ptr;
ptr = (int *) malloc(sizeof(*ptr));