In C,is casting to (void*) not needed/inadvisable for memcpy() just as it is not needed for malloc()?

I have some confusions about what I read from the following site about memcpy()(and malloc()):

http://www.cplusplus.com/reference/cstring/memcpy/

In that page,the following 2 lines are clearly stated:

destination

Pointer to the destination array where the content is to be copied, type-casted to a pointer of type void*.

source

Pointer to the source of data to be copied, type-casted to a pointer of type const void*.

But immediately after that,in the code,there is no casting to void* in the following two lines where memcpy() is used:

    memcpy ( person.name, myname, strlen(myname)+1 );
    memcpy ( &person_copy, &person, sizeof(person) );

Please answer the following 2 questions arising from this premise:

1) In C's case(as opposed to C++) is it all right and advisable not to cast to void* the return type or the arguments in memcpy() just as it is all right and advisable not to cast to void* the return type of malloc() in C?If so,as I intuitively feel, why is it explicitly stated in that reputed site that we need to cast it to void* (even though it doesn't use that in the code).Is that site wrong about it?

2) Now the real contradiction about that reputed site.Consider the following

http://www.cplusplus.com/reference/cstdlib/malloc/

In the case of malloc() ,in the description, it is written as if it is optional to cast to void* the return type (exact words "..can be cast to the desired type.."),unlike in the case of memcpy() above where it is said that it is to be cast into void*.But while in memcpy() the casting is not done even though it is written that it is to be cast,in the case of malloc(),the casting to void* is done even though it's written it can be cast to void*.Now I see something wrong in this as for C we are not supposed to cast malloc()'s return to void*.

To put the discrepancies in a nutshell again lest the people answering get confused in my wordy description:

--Is it advisable in C not to cast to void* the return and arguments of memcpy()?

--Is that site wrong about malloc() as it casts malloc() return to void* in C code.

From ISO/IEC 9899:2011 of C language specification, section 6.3.2.3, page 55:

A pointer to void may be converted to or from a pointer to any object type. A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.

So you basically never need to cast the result of a void* to the desired type, neither you need to do the opposite.