2 bytes allocated problems

[Matthias Andree]

Am 24.02.2010, 20:55 Uhr, schrieb Dag-Erling Smørgrav <des at 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