Is a conversion to `void *` required before converting a pointer to `uintptr_t` and vice versa?

Section 7.18.1.4 of the C99 standard states:

The following type designates an unsigned integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer:

uintptr_t

Does this mean that only void * types can be converted to uintptr_t and back without changing the value of the original pointer?

In particular, I would like to know if the following code is required to use uintptr_t:

int foo = 42;
void * bar = &foo;
uintptr_t baz = bar;
void * qux = baz;
int quux = *(int *)qux; /* quux == foo == 42 */

Or if this simpler version is guaranteed by the C99 standard to result in the same effect:

int foo = 42;
uintptr_t bar = &foo;
int baz = *(int *)bar; /* baz == foo == 42 */

Is a conversion to void * required before converting a pointer to uintptr_t and vice versa?