C++11: reference-to-void?

Question

Is this C++11 program ill-formed?

typedef void& rv;

int main() {}

I couldn't find anything disallowing it in the standard (looked in 3.9.2 and 8.3.2).

Clang says "cannot form a reference to 'void'", gcc says "cannot declare reference to ‘void’"

I would have expected [dcl.ref]/5 to give such a restriction, if it were intended.

Are the implementations just "reading between the lines" because such a type can never be used in an object definition?

Answer 1

Quoting from the C++11 standard (emphasis mine):

8.3.2 References

¹ [...] A declarator that specifies the type “reference to cv void” is ill-formed.

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As a side tidbit of info, this rule has been taken advantage by the N3421 greater<> proposal (which was already accepted to be included into the future C++14 standard) by having void as the default template argument.

B. The technique of using default template arguments and explicit specializations for void was chosen for its non-intrusiveness. greater<void> isn't valid C++11 (it would attempt to form a reference to void, forbidden by 8.3.2 [dcl.ref]/1). Additionally, while users are permitted to specialize Standard Library machinery (17.6.4.2.1 [namespace.std]/1), such specializations must involve user-defined types.

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Maybe I'm just a bit more indulged into these mind games, but for the rationale of why this rule is enforced -- and I must state that this is just my own humble opinion -- consider the following code:

sizeof(double&);   // Results in double's size

using foo_ref = foo&;
sizeof(foo_ref);   // Results in foo's size

sizeof(void);      // Error! void has no size!
sizeof(void&);     // Error! Tries to get the size of void

Most importantly, keep in mind that references are just aliases of other objects, and you cannot have an object of type void.