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Message-ID: <47F63641.4090103@knosof.co.uk>
Date: Fri, 04 Apr 2008 15:08:01 +0100
From: Derek M Jones <derek@...sof.co.uk>
To: Al Viro <viro@...IV.linux.org.uk>
CC: Josh Triplett <josh@...edesktop.org>,
Linus Torvalds <torvalds@...ux-foundation.org>,
Harvey Harrison <harvey.harrison@...il.com>,
Andrew Morton <akpm@...ux-foundation.org>,
LKML <linux-kernel@...r.kernel.org>, linux-sparse@...r.kernel.org
Subject: Re: [PATCH 7/7] asm-generic: suppress sparse warning in ioctl.h
Al,
> * typedef *can* be variably-modified, but only in block scope.
> Warning: this can get sticky for us - all sizes are evaluated when
> typedef is reached. IOW,
> typedef int a[n];
> a x;
> if (n++ == 5) {
> a y;
> int z[n];
> }
> will have size of y equal to that of x, but *not* equal to that of z.
An opportunity for Sparse to issue a warning :-)
> Another thing to keep in mind is that sizeof(VLA) is not a constant
> expression *and* that sizeof argument is evaluated. IOW, not only
> sizeof(int [n = f()])
> has side effects, but in
> int (*p)[n];
> ...
> sizeof(*(g(), p))
> will have g() evaluated. Note that if p had been declared as
> int (*p)[4];
> the same sizeof() would *NOT* have called anything. This, BTW, is where
> the rules for what an integer constant expression is are getting bloody
> important:
Any side effect appearing in a sizeof operand ought to be flagged.
There are people out there who think that the side effects occur (even
in C90).
Sentence 1122: http://c0x.coding-guidelines.com/6.5.3.4.html
"If the type of the operand is a variable length array type, the operand
is evaluated;"
But watch out for sentence 1584:
http://c0x.coding-guidelines.com/6.7.5.2.html
"Where a size expression is part of the operand of a sizeof operator and
changing the value of the size expression would not affect the result of
the operator, it is unspecified whether or not the size expression is
evaluated."
> int n = 1;
> int f(void)
> {
> int (*p)[1 + (0 && n)];
> return sizeof(*(g(), p));
> }
>
> _must_ call g() according to C99, while the same with
> enum {n = 1;} must not, even though n won't be even looked at in either
> case *and* any sane compiler will find return value of f() at compile
> time, turning it to
Any sane compiler that performs the analysis needed to deduce that
the expression inside the parenthesis always evaluated to zero
will turn it into....
> One more thing: use of sizeof(variably-modified type) may or may not
> evaluate size expressions in there if they do not affect the result.
> IOW, it's unspecified whether
> sizeof(int (*)[n++])
> increments n; different compilers are broken in different ways and it
This language feature came about because at least one vendor on the
WG14 committee had a compiler that optimized away subexpressions
within a sizeof that did not contribute to the result of the
evaluation. My attempt to stop the behavior being unspecified
did not succeed :-(
--
Derek M. Jones tel: +44 (0) 1252 520 667
Knowledge Software Ltd mailto:derek@...sof.co.uk
Applications Standards Conformance Testing http://www.knosof.co.uk
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