[cfe-dev] sizeof(sizeof(...))

[Bill Wendling]

Well, the standard says to "integer" type, so probably not unsigned.
But that's my understanding of it. Though I don't believe that sizeof
should be evaluating the expression...

-bw

On Dec 14, 2007 1:11 PM, Ted Kremenek <kremenek at apple.com> wrote:
> After thinking about this a little more, my understanding is that
> sizeof simply evaluates to a type of unsigned int, so
> sizeof(sizeof(...)) is equivalent to sizeof(unsigned int).  If anyone
> thinks this is wrong, please let me know.
>
>
> On Dec 14, 2007, at 12:30 PM, Ted Kremenek wrote:
>
> > Does anyone have any insights on the semantics of expressions of the
> > form sizeof(sizeof(...))?  For example, the following is legal code:
> >
> > int baz(int x) {
> >  typedef int a[f()];
> >  return sizeof (sizeof(a[bar(x)]));
> >         + sizeof(sizeof(x));
> > }
> >
> > From the C99 standard (6.5.3.4):
> >
> >   "The sizeof operator yields the size (in bytes) of its operand,
> > which may be an
> >   expression or the parenthesized name of a type. The size is
> > determined from the type of
> >   the operand. The result is an integer. Ifthe type of the operand is
> > a variable length array
> >   type, the operand is evaluated; otherwise, the operand is not
> > evaluated and the result is an
> >   integer constant."
> >
> > I am having a little difficulty interpreting the (general) semantics
> > of applying sizeof to sizeof.  Moreover, "sizeof (sizeof(a[bar(x)]))"
> > does not cause bar() to be called, whereas (as expected)
> > sizeof(a[bar(x)]) does.
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