Question

In this code, why is sizeof(x) the size of a pointer, not the size of the type x?

typedef struct {
  ...
} x;

void foo() {
  x *x = malloc(sizeof(x));
}

Answer 1

Because C says:

(C99, 6.2.1p7) “Any other identifier has scope that begins just after the completion of its declarator.”

So in your example, the scope of the object x start right after the x *x:

x *x = /* scope of object x starts here */
       malloc(sizeof(x));

To convince yourself, put another object declaration of type x right after the declaration of the object x: you will get a compilation error:

void foo(void)
{
    x *x = malloc(sizeof(x));  // OK
    x *a;   // Error, x is now the name of an object
}

Otherwise, as Shahbaz notee in the comments of another answer, this is still not a correct use of malloc. You should call malloc like this:

T *a = malloc(sizeof *a);

and not

T *a = malloc(sizeof a);