In article <6t3tan$57k$1@mulga.cs.mu.OZ.AU>,
  fjh@cs.mu.OZ.AU (Fergus Henderson) wrote:
> Edward Diener <eddielee@abraxis.com> writes:
>
> >Is the expression "sizeof( long [] )" legal in Ansi C++ as opposed to
> >"sizeof(long *)" which I know is legal ?
>
> No, `long []' is an incomplete type, and the type used in a sizeof()
> expression must be a complete type (see 5.3.3 paragraph 1).

Could you elaborate on this?  I've got a similar problem (can't sizeof() a
perfectly good struct.  What is this 5.3.3 paragraph 1?

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 <eanderso@usa.net> wrote:
>
>In article <6t3tan$57k$1@mulga.cs.mu.OZ.AU>,
>  fjh@cs.mu.OZ.AU (Fergus Henderson) wrote:
>> Edward Diener <eddielee@abraxis.com> writes:
>>
>> >Is the expression "sizeof( long [] )" legal in Ansi C++ as opposed to
>> >"sizeof(long *)" which I know is legal ?
>>
>> No, `long []' is an incomplete type, and the type used in a sizeof()
>> expression must be a complete type (see 5.3.3 paragraph 1).
>
>Could you elaborate on this?  I've got a similar problem (can't sizeof() a
>perfectly good struct.  What is this 5.3.3 paragraph 1?

   -1- The sizeof operator yields the number of bytes in the object
   representation of its operand. The operand is either an expression,
   which is not evaluated, or a parenthesized type-id. The sizeof
   operator shall not be applied to an expression that has function or
   incomplete type, or to an enumeration type before all its enumerators
   have been declared, or to the parenthesized name of such types, or to
   an lvalue that designates a bit-field. sizeof(char), sizeof(signed
   char) and sizeof(unsigned char) are 1; the result of sizeof applied to
   any other fundamental type (basic.fundamental) is implementation-
   defined.

A struct is "complete" only when its definition has been seen:

  struct a;  // incomplete
  int b[sizeof(a)];  // error, a incomplete
  struct a {};  // now complete
  int c[sizeof(a)];  // OK.

--
Nathan Myers
ncm@nospam.cantrip.org  http://www.cantrip.org/



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Edward Diener <eddielee@abraxis.com> writes:

>Is the expression "sizeof( long [] )" legal in Ansi C++ as opposed to
>"sizeof(long *)" which I know is legal ?

No, `long []' is an incomplete type, and the type used in a sizeof()
expression must be a complete type (see 5.3.3 paragraph 1).

--
Fergus Henderson <fjh@cs.mu.oz.au>  |  "I have always known that the pursuit
WWW: <http://www.cs.mu.oz.au/~fjh>  |  of excellence is a lethal habit"
PGP: finger fjh@128.250.37.3        |     -- the last words of T. S. Garp.
---
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Is the expression "sizeof( long [] )" legal in Ansi C++ as opposed to
"sizeof(long *)" which I know is legal ? The MIDL compiler in Windows
produces the former within code on same occasions given a valid IDL
construct .



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