Why need the (void *) in the pointer conversion?

Question

In the following code, why need the (VOID *) in the middle of pointer conversion?

Some Context definition:

#define VOID      void
typedef unsigned char       UINT8;


ErrorDest = (EFI_ACPI_6_0_GENERIC_HARDWARE_ERROR_SOURCE_STRUCTURE *)
            (VOID *)
            ((UINT8 *)Hest + Hest->Header.Length);

Answer 1

The only reason I can see is to prevent the compiler from giving warnings about pointer alignment.

UINT8 has no alignment requirements.

A struct can have other alignment requirements depending on its members. Ususally the alignment requirement of a struct is the same as the largest alignment requirement of its members. If EFI_ACPI_6_0_GENERIC_HARDWARE_ERROR_SOURCE_STRUCTURE is an opaque type (that is, it is declared, but its members are unknown), or if it contains members with alignment requirements then the compiler may issue a warning about the alignment.

A void* may point to any adress but is excempt from alignment warnings when it is cast to another pointer type, even if that type may have alignment requirements.

Answer 2

The reason might be alignment, as pointed out in another answer. But it might as well just be a bug - a failed attempt to dodge incompatible pointer type errors.

Upon writing code such as this_t* a = (that_t*)b; you get a compiler error about incompatible types. It is then a somewhat common mistake to attempt to fix this error by writing

this_t* a = (void*)(that_t*)b; // bad, dont do this

Indeed this silences the compiler error. And instead of a compiler error, we now have a run-time undefined behavior bug, because of strict aliasing violation.

If EFI_ACPI_6_0_GENERIC_HARDWARE_ERROR_SOURCE_STRUCTURE doesn't contain a member which is an uint8_t [], then the code in the question invokes undefined behavior due to strict aliasing violation.