Date: Thu, 25 Feb 2010 23:07:15 +0100 From: "Matthias Andree" <matthias.andree@gmx.de> To: =?utf-8?Q?Dag-Erling_Sm=C3=B8rgrav?= <des@des.no>, "Andrey Zonov" <andrey.zonov@gmail.com> Cc: freebsd-hackers@freebsd.org Subject: Re: 2 bytes allocated problems Message-ID: <op.u8peedxy1e62zd@merlin.emma.line.org> In-Reply-To: <86sk8qpfx8.fsf@ds4.des.no> References: <983a1cf21002240544s59006035ifbf0ef7eb045e44f@mail.gmail.com> <86eikar7gv.fsf@ds4.des.no> <4B858007.1000008@gmail.com> <86sk8qpfx8.fsf@ds4.des.no>
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Am 24.02.2010, 20:55 Uhr, schrieb Dag-Erling Smørgrav <des@des.no>:
> Why is there a 0 after the 'i'? Because when you write "abcdefghi", the
> compiler actually stores "abcdefghi\0". That's the definition of
> "string" in C: a sequence of characters immediately followed by a 0. If
> you don't want the 0 there, you have to do something like this:
>
> char a[9] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i' };
char a[9] = "abcdefghi";
suffices. The compiler knows there isn't room for the terminal '\0' and
omits it.
char a[] = "abcdefghi";
would append the implicit \0 and reserve 10 bytes for a.
> but then you don't have a string, just an array of 9 chars.
Not that the compiler itself could/would tell the difference after
initialization, or that it would even care. It's the library functions
that expect strings that care about the \0.
Beyond that, I recommend a C book to Andrey.
--
Matthias Andree
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