Guennadi Liakhovetski wrote:
quoted text
> On Thu, 2 Aug 2007, Andi Kleen wrote:
> 
>> Guennadi Liakhovetski <g.liakhovetski@gmx.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.

quoted text
> 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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