array initialization

[arne.muller@gmail.com]

Hello,

I was wondering if the following statement will initialize all the 6
elements with NULL:

void function() {
int *list[6] = {NULL};
. ..
}

what I want is an array with 6 pointers to int that are initialized
with NULL. Or do I've to do

int *list[6] = {NULL, NULL, NULL, NULL, NULL, NULL} ?

thanks for your help,
+kind regards,

Arne

[pete]

Re: array initialization

arne.muller@gma il.com wrote:

>
Hello,
>
I was wondering if the following statement will initialize all the 6
elements with NULL:
>
void function() {
int *list[6] = {NULL};
. ..
}

Yes, it will.
A short initializer for an array,
means that all of the elements
following the initialized one or initialized ones,
are initialized with a value of zero,
and zero means the same thing as NULL
when used to initialize a pointer.

what I want is an array with 6 pointers to int that are initialized
with NULL. Or do I've to do
>
int *list[6] = {NULL, NULL, NULL, NULL, NULL, NULL} ?

No, you don't.

--
pete

[assiss]

Re: array initialization

int *list[6] = {NULL};
list[0]=NULL, then the other 5 elements will be initialized with 0.

ONLY if you are sure that NULL equals to 0.

Thu, 24 Aug 2006 15:49:10 -0700，arne.mul ler wrote:

Hello,
>
I was wondering if the following statement will initialize all the 6
elements with NULL:
>
void function() {
int *list[6] = {NULL};
. ..
}
>
what I want is an array with 6 pointers to int that are initialized
with NULL. Or do I've to do
>
int *list[6] = {NULL, NULL, NULL, NULL, NULL, NULL} ?
>
thanks for your help,
+kind regards,
>
Arne

[Eric Sosman]

Re: array initialization

assiss wrote:

int *list[6] = {NULL};
list[0]=NULL, then the other 5 elements will be initialized with 0.
>
ONLY if you are sure that NULL equals to 0.

I am sure that NULL equals 0. Always. On every
Standard-conforming C implementation ever built. Amen.

--
Eric Sosman
esosman@acm-dot-org.invalid

[Robert Gamble]

Re: array initialization

Eric Sosman wrote:

assiss wrote:
>

int *list[6] = {NULL};
list[0]=NULL, then the other 5 elements will be initialized with 0.

ONLY if you are sure that NULL equals to 0.

>
I am sure that NULL equals 0. Always. On every
Standard-conforming C implementation ever built. Amen.

assiss may be confusing the *value* of a null pointer constant with the
*representation * of a null pointer constant. He would be well-advised
to re-read Chapter 5 of the FAQ <http://c-faq.com/>.

Robert Gamble

[assiss]

Re: array initialization

Robert Gamble wrote:

Eric Sosman wrote:

>assiss wrote:
>>

>>int *list[6] = {NULL};
>>list[0]=NULL, then the other 5 elements will be initialized with 0.
>>>
>>ONLY if you are sure that NULL equals to 0.

> I am sure that NULL equals 0. Always. On every
>Standard-conforming C implementation ever built. Amen.

>
assiss may be confusing the *value* of a null pointer constant with the
*representation * of a null pointer constant. He would be well-advised
to re-read Chapter 5 of the FAQ <http://c-faq.com/>.
>
Robert Gamble
>

thanks. I do forget someting about NULL.

[CBFalconer]

Re: array initialization

Eric Sosman wrote:

assiss wrote:
>

>int *list[6] = {NULL};
>list[0]=NULL, then the other 5 elements will be initialized with 0.
>>
>ONLY if you are sure that NULL equals to 0.

>
I am sure that NULL equals 0. Always. On every
Standard-conforming C implementation ever built. Amen.

Oh? Consider:

#include <stdio.h>

int main(void);
int n = 0;

if (n == NULL) puts('You are right');
else puts('You are wrong');
return 0;
}

Note that I did NOT write "if (0 == NULL)"

--
Chuck F (cbfalconer@yah oo.com) (cbfalconer@mai neline.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.netUSE maineline address!

[arne.muller@gmail.com]

Re: array initialization

Thanks for your replies! Fromù what I read in the FAQ one can use 0 or
NULL (macro) - but the little prog below just gives me a segmentation
fault ;-(

kind regards,

Arne

CBFalconer wrote:

Eric Sosman wrote:

assiss wrote:

int *list[6] = {NULL};
list[0]=NULL, then the other 5 elements will be initialized with 0.
>
ONLY if you are sure that NULL equals to 0.

I am sure that NULL equals 0. Always. On every
Standard-conforming C implementation ever built. Amen.

>
Oh? Consider:
>
#include <stdio.h>
>
int main(void);
int n = 0;
>
if (n == NULL) puts('You are right');
else puts('You are wrong');
return 0;
}
>
Note that I did NOT write "if (0 == NULL)"
>
--
Chuck F (cbfalconer@yah oo.com) (cbfalconer@mai neline.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.netUSE maineline address!

[Eric Sosman]

Re: array initialization

CBFalconer wrote:

Eric Sosman wrote:
>

>>assiss wrote:
>>
>>

>>>int *list[6] = {NULL};
>>>list[0]=NULL, then the other 5 elements will be initialized with 0.
>>>
>>>ONLY if you are sure that NULL equals to 0.

>>
> I am sure that NULL equals 0. Always. On every
>>Standard-conforming C implementation ever built. Amen.

>
>
Oh? Consider:
>
#include <stdio.h>
>
int main(void);
int n = 0;
>
if (n == NULL) puts('You are right');
else puts('You are wrong');
return 0;
}
>
Note that I did NOT write "if (0 == NULL)"

... and I for my part did NOT write "NULL equals zero."

(Also, it looks like your double-quote key is functioning
erratically: Sometimes it omits the left or right half -- hard
to tell which -- of its symbol. Have it checked before it gets
even worse, and maybe catches fire. ;-)

Back to the thread: assiss, Chuck and I have been having a
little fun at your expense, and I apologize. Here's the scoop:
Every variable that's "initialize d in part" is "initialize d in
full." If you provide an initializer for just a few elements
of an array or struct, all the other elements are initialized
for you. The elements you don't initialize explicitly are all
initialized to "zeroes of the proper type:" 0.0 for double,
'\0' for char, (char*)0 for char* pointers, and so on. (Union
objects are a little different because they can hold only one
value at a time, so when a union is initialized this way what
happens is that its first element is given a zero of the type
appropriate to that element.) This all works even on machines
where the various kinds of zero and NULL might not be represented
as all-bits-zero; it's the compiler's business to make it work.

What variables are initialized this way? Two kinds: those
with static storage duration (variables outside functions plus
`static' variables inside functions), and any variables for
which you provide a partial initializer:

int a; /* a == 0 */
int b = 3; /* b == 3 */
int c[4] = {1,2}; /* c == 1,2,0,0 */
struct s {
int x, y;
} d = { 1 }; /* d.x == 1, d.y == 0 */
union u {
char *cp;
int i;
float f;
} e; /* e.cp == (char*)0 */

void f(void) {
static int fa; /* fa == 0 */
int fc[4] = {9}; /* fc == 9,0,0,0 */
struct s fd = { 2 }; /* fd.x == 2, fd.y == 0 */
int v; /* NOT INITIALIZED */
}

The newer "C99" version of the Standard allows some fancier
forms of initializers (for example, you can provide explicit
initializers for elements [5],[1],[9] of an array, in that order),
but even then the rule holds: Initialize any part of something,
and all the parts you don't initialize receive zeroes.

--
Eric Sosman
esosman@acm-dot-org.invalid

[assiss]

Re: array initialization

Eric Sosman wrote:

CBFalconer wrote:

>Eric Sosman wrote:
>>

>>assiss wrote:
>>>
>>>
>>>int *list[6] = {NULL};
>>>list[0]=NULL, then the other 5 elements will be initialized with 0.
>>>>
>>>ONLY if you are sure that NULL equals to 0.
>>>
>> I am sure that NULL equals 0. Always. On every
>>Standard-conforming C implementation ever built. Amen.

>>
>>
>Oh? Consider:
>>
>#include <stdio.h>
>>
>int main(void);
> int n = 0;
>>
> if (n == NULL) puts('You are right');
> else puts('You are wrong');
> return 0;
>}
>>
>Note that I did NOT write "if (0 == NULL)"

>
... and I for my part did NOT write "NULL equals zero."
>
(Also, it looks like your double-quote key is functioning
erratically: Sometimes it omits the left or right half -- hard
to tell which -- of its symbol. Have it checked before it gets
even worse, and maybe catches fire. ;-)
>
Back to the thread: assiss, Chuck and I have been having a
little fun at your expense, and I apologize. Here's the scoop:
Every variable that's "initialize d in part" is "initialize d in
full." If you provide an initializer for just a few elements
of an array or struct, all the other elements are initialized
for you. The elements you don't initialize explicitly are all
initialized to "zeroes of the proper type:" 0.0 for double,
'\0' for char, (char*)0 for char* pointers, and so on. (Union
objects are a little different because they can hold only one
value at a time, so when a union is initialized this way what
happens is that its first element is given a zero of the type
appropriate to that element.) This all works even on machines
where the various kinds of zero and NULL might not be represented
as all-bits-zero; it's the compiler's business to make it work.
>
What variables are initialized this way? Two kinds: those
with static storage duration (variables outside functions plus
`static' variables inside functions), and any variables for
which you provide a partial initializer:
>
int a; /* a == 0 */
int b = 3; /* b == 3 */
int c[4] = {1,2}; /* c == 1,2,0,0 */
struct s {
int x, y;
} d = { 1 }; /* d.x == 1, d.y == 0 */
union u {
char *cp;
int i;
float f;
} e; /* e.cp == (char*)0 */
>
void f(void) {
static int fa; /* fa == 0 */
int fc[4] = {9}; /* fc == 9,0,0,0 */
struct s fd = { 2 }; /* fd.x == 2, fd.y == 0 */
int v; /* NOT INITIALIZED */
}
>
The newer "C99" version of the Standard allows some fancier
forms of initializers (for example, you can provide explicit
initializers for elements [5],[1],[9] of an array, in that order),
but even then the rule holds: Initialize any part of something,
and all the parts you don't initialize receive zeroes.
>

Thanks very much.
I check out ansi_c99 and find that I have some wrong understandings
about initializations .

[CBFalconer]

Re: array initialization

arne.muller@gma il.com wrote:

>
Thanks for your replies! Fromù what I read in the FAQ one can use
0 or NULL (macro) - but the little prog below just gives me a
segmentation fault ;-(

Please do not top-post. Some details and reasons in the links
below.

--
Some informative links:
news:news.annou nce.newusers

http://www.geocities.com/nnqweb/

http://www.catb.org/~esr/faqs/smart-questions.html

http://www.caliburn.nl/topposting.html

http://www.netmeister.org/news/learn2quote.html

[Frederick Gotham]

Re: array initialization

Eric Sosman posted:

union u {
char *cp;
int i;
float f;
} e; /* e.cp == (char*)0 */

Are you saying that zero-intialisation for a union-type results in zero-
initialisation of its first member?

--

Frederick Gotham

[Robert Gamble]

Re: array initialization

Frederick Gotham wrote:

Eric Sosman posted:
>

union u {
char *cp;
int i;
float f;
} e; /* e.cp == (char*)0 */

>
>
Are you saying that zero-intialisation for a union-type results in zero-
initialisation of its first member?

I think he is saying that a static variable of union type has it's
first member initialized to zero which is correct.

Robert Gamble

[Eric Sosman]

Re: array initialization



Frederick Gotham wrote On 08/25/06 10:16,:

Eric Sosman posted:
>
>

> union u {
> char *cp;
> int i;
> float f;
> } e; /* e.cp == (char*)0 */

>
>
>
Are you saying that zero-intialisation for a union-type results in zero-
initialisation of its first member?

Give the man a cigar! (Section 6.7.8, paragraph 10.)

Notes: (1) The example above came from a larger example
in which it was clear that the declaration was at file scope,
hence had static storage duration, and hence was initialized
even without an explicit initializer. (2) It's actually not
the union's first element that gets initialized, but its
first *named* element.

union u2 {
int : 4; /* unnamed bit-field */
int : 13; /* another one */
char *cp; /* first named element */
double d;
} f; /* f.cp == (char*)0 */

Unnamed elements in unions seem pretty useless, but as far
as I can see they're permitted.

--
Eric.Sosman@sun .com

[Flash Gordon]

Re: array initialization

assiss wrote:

Your reply belongs under (or interspersed with) the text you are
replying to. NEVER above. I've corrected it this time.

Thu, 24 Aug 2006 15:49:10 -0700，arne.mul ler wrote:
>

>Hello,
>>
>I was wondering if the following statement will initialize all the 6
>elements with NULL:
>>
>void function() {
> int *list[6] = {NULL};
> . ..
>}

<snip>

int *list[6] = {NULL};
list[0]=NULL, then the other 5 elements will be initialized with 0.
>
ONLY if you are sure that NULL equals to 0.

Wrong. The C standard defines that the other five elements will be
initialised to null pointers (not NULL which is a macro name). This will
be the case however null pointers are represented.
--
Flash Gordon.