Re: array size 1 ? All headers

["Luiz Fernando N. Capitulino" <lcapitulino@terra.com.br>]

 Hi Gre,

Em Wed, 21 Jul 2004 10:27:35 +0800
"Gre Taguran" <gtaguran@sni.ph> screveu:

| i think its not array of 1 but instead array of 2 which is 0 and 1 index.

 Wrong.

 K&R defines array as: array[lenght]

 So, if you defines an array as:

 int luiz[10]

 You will have 10 memory cells for it, and if you do:

 int luiz[1]

 You will have one memory cell.

 But, the _access_ is made from 0, so for the first example, we have:

 luiz[0], luiz[1], luiz[2]... luiz[9].

 For the second, just:

 luiz[0].

| maybe 
| it is because u were thinking that the last array index (in this case is 1) is 
| null, but it is only applicable when handling string because string needs a 
| terminated string. ex.
| 
| char tst[1];
| tst[0]='g';
| tst[1]='r';
| printf("%s",tst); //this would print gr<and more chars here because it is not 
| terminated
| printf("%c=%c"); //this will print gr meaning tst[1] is alocated by 'r'

 The program:

 char tst[1];
 tst[0]='g';
 tst[1]='r';

 printf("%c %c\n", tst[0], tst[1]);

 compiles and works here! but to be honest, I don't why. I can prove it is
an odd case, becase this also works (here):

 char tst[1];
 tst[0]='g';
 tst[1]='r';
 tst[2]='z';

 printf("%c %c %c\n", tst[0], tst[1], tst[2]);

PS: GCC compiled.

-- 
Luiz Fernando