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Message-id: <46B24FC5.8060409@shaw.ca>
Date: Thu, 02 Aug 2007 15:42:29 -0600
From: Robert Hancock <hancockr@...w.ca>
To: Guennadi Liakhovetski <g.liakhovetski@....de>
Cc: Andi Kleen <andi@...stfloor.org>, linux-kernel@...r.kernel.org
Subject: Re: gcc fixed size char array initialization bug - known?
Guennadi Liakhovetski wrote:
> On Thu, 2 Aug 2007, Andi Kleen wrote:
>
>> Guennadi Liakhovetski <g.liakhovetski@....de> writes:
>>
>>> char c[4] = "0123";
>>> and - a wonder - no warning.
>> It's required by the C standard.
>>
>> 6.7.8.14 of C99:
>> ``
>> An array of character type may be initialized by a character string literal, optionally
>> enclosed in braces. Successive characters of the character string literal (including the
>> terminating null character if there is room or if the array is of unknown size) initialize the
>> elements of the array.
>> ''
>>
>> Note the "if there is room".
>>
>> I believe the rationale is that it still allows to conveniently initialize
>> non zero terminated strings.
>
> Right, I accept that it will compile, but I don't understand why "01234"
> produces a warning and "0123" doesn't? Don't think C99 says anything about
Because 5 characters will not fit in a 4 character array, even without
the null terminator.
> that. And, AFAIU, using structs with fixed-size char array we more or less
> rely on the compiler warning us if anyone initializes it with too long a
> string.
>
> Also interesting, that with
>
> char c[4] = "012345";
>
> the compiler warns, but actually allocates a 6-byte long array...
>
> Thanks
> Guennadi
--
Robert Hancock Saskatoon, SK, Canada
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Home Page: http://www.roberthancock.com/
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