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

[Stefan Richter <stefanr@s5r6.in-berlin.de>]

Al Viro wrote:
> 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.

No, the fact that the full 012345\0 ends up in the object file is
apparently unrelated to what happens to the variable c...

> 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.

...since only 0123 will get into c at runtime, i.e. a 4 bytes long array
without \0 appendix or other extraordinary padding.

#include <stdio.h>
#include <string.h>

int main()
{
	char c[4] = "012345";

	printf("%d %d _%s_\n", sizeof c / sizeof *c, strlen(c), c);
	return 0;
}

$ ./a.out
4 8 _01230®¿_

$ strings a.out |grep 0123
012345

-- 
Stefan Richter
-=====-=-=== =--- ---==
http://arcgraph.de/sr/