How is void *a = &a legal?

Question

Consider the following C++ code:

void* a = &a;

Why doesn't the compiler complain for using an undeclared identifier?

Also, what does the compiler consider the variable a to be? Is it a pointer to a void object or is it a pointer to a void* pointer?

[Point of Declaration](http://msdn.microsoft.com/en-us/library/179x7xb9.aspx) — Grijesh Chauhan, Aug 14 '13 at 07:32
[Point of Declaration in C++](http://stackoverflow.com/a/15746272/952747) — masoud, Aug 14 '13 at 08:01
Should mention why you'd want to do this - to get a pointer to the top of the stack (from which you can fiddle with all sorts of things). — OrangeDog, Aug 14 '13 at 08:50

score 95 · Accepted Answer · answered Aug 14 '13 at 06:40

95

The scope of declaration of variables in C++ can be pretty surprising:

void* a =               &a;
         ^~~~~~~~~~~~~~~~~
          a declared as `void*` from here on

Therefore, &a is void** but since any pointer type is implicitly convertible to void*...

answered Aug 14 '13 at 06:40

Matthieu M.

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How would I assign a useful object to this? – user2681063 Aug 14 '13 at 06:53
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@user2681063 `a = &userfulObject`? – Nikos C. Aug 14 '13 at 06:59
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@MarkGarcia: Note that `void *a = a;` would be UB if declared locally, otherwise it is fine at namespace scope. – Nawaz Aug 14 '13 at 07:01
Is this also applicable in C? – Theodoros Chatzigiannakis Aug 14 '13 at 11:26
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@TheodorosChatzigiannakis: I would believe it is, yes. – Matthieu M. Aug 14 '13 at 12:38
Ok, now the question becomes: **WHY** is this allowed? I understand scoping, but I don't understand why you would want to do `int a; void *a = &a`? – Cole Tobin Aug 20 '13 at 21:23
@Cole"Cole9"Johnson: there is no talk of `int a;` in my demonstration so I am not sure where you are going on... regardless, if we simply stick to the example, I would say it is allowed because there is no reason **not to** and it would be extra effort to forbid it. – Matthieu M. Aug 21 '13 at 06:07
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A spec citation would be really nice here. – Benjamin Gruenbaum Aug 21 '13 at 11:18
@BenjaminGruenbaum: I can see 3 points for citation (declaration, address of uninitialized variable and conversion to `void*`), but unfortunately finding either in the specs is always an uphill battle... – Matthieu M. Aug 21 '13 at 12:55
@MatthieuM. Doesn't mean it can't be done. I always use the ToC – Cole Tobin Aug 21 '13 at 14:32

score 30 · Answer 2 · edited Aug 14 '13 at 09:30

30

It is equivalent to

void* a;
a = &a;

Therefore, a has been declared. So a gets the address of a written in a. So it is a pointer to a void pointer. (You did not define any objects yet.)

edited Aug 14 '13 at 09:30

Ilmari Karonen

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answered Aug 14 '13 at 06:39

Stasik

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Technically, you did define one object. `a` itself is a object. (Not all objects have user-defined types in C++) – MSalters Aug 14 '13 at 07:10
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@MSalters what is this 'object' you speak of? :D – Gusdor Aug 14 '13 at 15:19
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@Gusdor We can only assume what lies beyond the void of the event horizon. – Cole Tobin Aug 20 '13 at 21:24
They are called differently, however the "effect" is the same in this case – Stasik Aug 23 '13 at 07:58
They are not equivalent in general, but in _this_ scenario they are. – Lightness Races in Orbit Mar 22 '14 at 17:23

Mario Rossi · Answer 3 · 2013-08-14T07:25:17.150

In void* a, a is declared as a pointer not to a void type but to "any" type (special case). An address (position in memory) is assigned to a, as to any other variable being declared, of course.

After that, expression &a is evaluated to initialize the variable (also a, but this is not relevant) just declared. The type of &a is "pointer to pointer to any type", which is a special case of "pointer to any type", fully compatible with the type of a. Ergo, no compiler message.

Corollary: do not use void* if you want strong type checking. Anything can be converted to it. Just the opposite in the reverse direction, except for void* itself (it would be an unnecessary exception that a type was incompatible with itself).

Also, AFAIR this really comes from C.

How is void *a = &a legal?

3 Answers3