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

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

Guennadi Liakhovetski wrote:
> On Thu, 2 Aug 2007, Andi Kleen wrote:
>> 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 

How should gcc know whether you actually wanted that char foo[len] to
contain a \0 as last element?

Given the respective command line switches, gcc does warn in some cases
where it is guesswork whether what you typed is what you intended.  For
example

	if (i = j())

is reason for gcc to warn even if that might exactly be what you wanted.
However this construct can easily be annotated as

	if ((i = j()))

to show to gcc and to carbon-based bipedals that you indeed wanted this.

Now there is no nice way to make an annotation that says "look, I'm
going to initialize an array of char with a string literal now, and the
resulting array will contain a non-zero member as last element, and I
mean it".  And since there is no such annotation possible, gcc does not
warn and demand that you annotate the perfectly valid and 100%
spec-compliant construct char a[4] = "1234";.
-- 
Stefan Richter
-=====-=-=== =--- ---==
http://arcgraph.de/sr/
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