The C standard defines the behavior of the free function:
The free function causes the space pointed to by ptr to be
deallocated, that is, made available for further allocation.
which means that a later call to malloc (or something else) might re-use the same memory space.
As soon as a pointer is passed to free(), the object it pointed to reaches the end of its lifetime. Any attempt to refer to the pointed-to object has undefined behavior (i.e., you're no longer allowed to dereference the pointer).
More than that, the value of the pointer itself becomes indeterminate, so any attempt to refer to the pointer value has undefined behavior. Reference: N1570 6.2.4p2:
If an object is referred to outside of its lifetime, the behavior is
undefined. The value of a pointer becomes indeterminate when the
object it points to (or just past) reaches the end of its lifetime.
It's true that free()'s argument is passed by value (like all C function arguments), and so free can't actually modify the pointer. One way to think of it is that the pointer has the "same" value before and after the call, but that value is valid before the call and indeterminate after the call.
It's likely that an attempt to refer to the pointer value, or even to dereference it, will appear to "work". That's one of the many possible symptoms of undefined behavior (and arguably the worst, since it makes it difficult to detect and diagnose the error).
