Why can't an incomplete type be casted to void?

Question

Why does the following code give the following error?

Why does the type need to be complete in order to be casted to void?

struct Incomp         


        

Answer 1

It's a change between C and C++, where Microsoft previously implemented the C rules. As noted in remyabel's answer, that has since been fixed.

In C, a cast to void, or simply using an expression as a statement by itself (as in incomplete;), still involves the lvalue-to-rvalue conversion. C calls it slightly differently, but it's the same conversion.

In C++, a cast to void, or simply using an expression as a statement by itself doesn't involve the lvalue-to-rvalue conversion. This is needed because C++ makes assignment operators return lvalues, so if the lvalue-to-rvalue conversion were applied, then

volatile int i;
i = 1;

would not merely store, it would also immediately load afterwards.

The lvalue-to-rvalue conversion requires a complete type, even if the value is then discarded, since otherwise, it's impossible to know how many bytes should be read.

Answer 2

I don't see anything prohibiting this, quoting N4140:

§5.4/4 The conversions performed by

[...]

— a static_cast (5.2.9),

[...]

can be performed using the cast notation of explicit type conversion.

§5.2.9/5 Otherwise, the static_cast shall perform one of the conversions listed below. No other conversion shall be performed explicitly using a static_cast.

§5.2.9/6 Any expression can be explicitly converted to type cv void, in which case it becomes a discarded-value expression (Clause 5). [...]

It's most likely a bug as tested on Rextester, an online VS2013 compiler, but compiles in rise4fun, Microsoft's online compiler which is bleeding edge.