Re: gcc fixed size char array initialization bug - known?

[Alexander van Heukelum]

On Fri, 3 Aug 2007 00:09:15 +0100, "Al Viro" <viro@xxxxxxxxxxxxxxxx>
said:
> On Fri, Aug 03, 2007 at 12:51:16AM +0200, Guennadi Liakhovetski wrote:
> > On Fri, 3 Aug 2007, Stefan Richter wrote:
> > 
> > > Guennadi Liakhovetski wrote:
> > > > with
> > > > 
> > > > 	char c[4] = "012345";
> > > > 
> > > > the compiler warns, but actually allocates a 6-byte long array...
> > > 
> > > Off-topic here, but:  sizeof c / sizeof *c == 4.
> > 
> > Don't think it is OT here - kernel depends on gcc. And, what I meant, is, 
> > that gcc places all 7 (sorry, not 6 as I said above) characters in the 
> > .rodata section of the compiled object file. Of course, it doesn't mean, 
> > that c is 7 characters long.
> 
> So gcc does that kind of recovery, after having warned you.  Makes sense,
> as long as it's for ordinary variables (and not, say it, struct fields) -
> you get less likely runtime breakage on the undefined behaviour (e.g.
> passing c to string functions).  So gcc has generated some padding
> between the global variables, that's all.
> 
> It doesn't change the fact that use of c[4] or strlen(c) or strcpy(...,
> c) means nasal demon country for you.
> 
> Now, if gcc does that for similar situation with struct fields, you'd
> have a cause to complain.

Hi!

(It took me a while before I understood that that last that referred to
padding inside a struct generated by gcc due to overlong initializers.)
But from the rest of the thread it seems that some people expect the
compiler to warn about the following...

struct {char c[4];} s1 = {"abcd"};

It doesn't. Of course if one wants to be warned in such cases
(initialisation
of a character array of specified length using a string constant) one
could
tell the compiler that the 0 at the end should really be there:

struct {char c[4];} s2 = {"abcd" "\0"};

Writing it like this will give them the expected warning.

Greetings,
    Alexander
-- 
  Alexander van Heukelum
  heukelum@xxxxxxxxxxx

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