Array Initialization problems.

[Trevor Qi]

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

Thanks!

Sincerely,
Trevor

[Victor Nazarov]

Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

As I know, there isn't, even in C99.

Explain any reaseon why you want to do it in two lines.

vir

[Mark A. Odell]

qh****@hotmail.com (Trevor Qi) wrote in
news:17**************************@posting.google.c om:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

Then no, you cannot. You could do something silly like:

x[0] = 0, x[1] = 0, x[2] = 1, x[3] = 2, x[4] = 9;

but that's probably not something that will scale too well. If x[] is
file-scoped you could create a function level array and the memcpy() it.
E.g.

#include <string.h>

int x[5];

int main(void)
{
int private_x[] = { 0, 0, 1, 2, 9 };

memcpy(x, private_x, sizeof x);

return 0;
}

--
- Mark ->
--

[Lew Pitcher]

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Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};
Not unless you use a typographical break rather than a statement break.

You'd have to do something like

int X[5]
= {0,0,1,2,9};

Otherwise, what you want to do is not possible, and your code sample is
incorrect.
But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

Thanks!

Sincerely,
Trevor

- --

Lew Pitcher, IT Consultant, Enterprise Application Architecture
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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[Tim Orbaker]

Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

It sounds to me like this is what you want (tested with gcc):

In a header named 'array.h', include the line:

extern int X[5];

Then, create another file (perhaps: array.c), include:

int X[5] = { 0, 0, 1, 2, 9 };

Then either #include 'array.h' in each file that uses this array or use:

extern int X[5];

in any lexical scope that uses the array.

Tim

[Emmanuel Delahaye]

In 'comp.lang.c', qh****@hotmail.com (Trevor Qi) wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

If it is a read-only data, you can define it

/* in a unique compile unit */
static int const X[] = {0,0,1,2,9};

or

/* shared constant */
int const X[];

with this declaration:
extern int const X[] = {0,0,1,2,9};

Example :

#include <stdio.h>

extern int const X[] = {0,0,1,2,9};

int const X[];

int main (void)
{
int i;

for (i = 0; i < sizeof X / sizeof *X; i++)
{
printf ("X[%d] = %d\n", i, X[i]);
}
return 0;
}

D:\CLC\T\TREVOR>bc
X[0] = 0
X[1] = 0
X[2] = 1
X[3] = 2
X[4] = 9

Or

/* const.c */
#include "const.h"

int const X[];

#ifndef H_ED_CONST_20040514223101
#define H_ED_CONST_20040514223101

/* const.h */
extern int const X[] = {0,0,1,2,9};

#endif /* guard */

/* Guards added by GUARD (c) ED 2000-2003 May 09 2004 Ver. 1.6 */

/* main.c */
#include <stdio.h>
#include "const.h"

int main (void)
{
int i;

for (i = 0; i < sizeof X / sizeof *X; i++)
{
printf ("X[%d] = %d\n", i, X[i]);
}
return 0;
}

--
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The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
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FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/

[Martin Ambuhl]

Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message. I know "int X[5] = {0,0,1,2,9};"
works.
I really want to do this because this is a public array and will be
used by many Subroutines.

Assignment of arrays is not part of C. To copy arrays at a go, use
memmove() or memcpy().

I have given another answer in the code below, but it will *not* work
with C89 compilers, which means it will not work with most actually
existing compilers. Even when it _does_ work, it may not be the best
approach.

int main(void)
{
int *X;
X = (int [5]){ 0, 0, 1, 2, 9}; /* make the pointer (not array) X
point to an anonymous array */
return 0;
}

[August Derleth]

On Fri, 14 May 2004 22:00:16 +0400, Victor Nazarov wrote:

Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};
As I know, there isn't, even in C99.

Then you know wrong: C99 permits compound literals.

Like so:

int *X;
X = (int [5]) {0,0,1,2,9};

Not exactly the same, but damned close.

Explain any reaseon why you want to do it in two lines.
A lot of other languages permit much the same thing.

vir

--
yvoregnevna gjragl-guerr gjb-gubhfnaq guerr ng lnubb qbg pbz
To email me, rot13 and convert spelled-out numbers to numeric form.
"Makes hackers smile" makes hackers smile.

[Ben Pfaff]

August Derleth <se*@sig.now> writes:

On Fri, 14 May 2004 22:00:16 +0400, Victor Nazarov wrote:

Trevor Qi wrote:

Hi,dear all:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

As I know, there isn't, even in C99.

Then you know wrong: C99 permits compound literals.

Like so:

int *X;
X = (int [5]) {0,0,1,2,9};

Not exactly the same, but damned close.

Even closer:
int X[5];
memcpy(X, (const int[5]) {0, 0, 1, 2, 9}, sizeof X);
--
"I hope, some day, to learn to read.
It seems to be even harder than writing."
--Richard Heathfield

[Old Wolf]

Martin Ambuhl <ma*****@earthlink.net> wrote:

Trevor Qi wrote:

I have the following question. If I want to initialize an array using
braces, is there a way that I can define them and give the initial
value in different lines, like:

int X[5];
X[5] = {0,0,1,2,9};

But this gives me error message.

Assignment of arrays is not part of C. To copy arrays at a go, use
memmove() or memcpy().

int *X;
X = (int [5]){ 0, 0, 1, 2, 9};

Is it completely portable to go:

int X[5];
memcpy(X, (int []){ 0, 0, 1, 2, 9}, 5 * sizeof(int));

(I imagine that a trap representation might be generated after copying
some of the bytes of an int)

[Ben Pfaff]

ol*****@inspire.net.nz (Old Wolf) writes:

Is it completely portable to go:

int X[5];
memcpy(X, (int []){ 0, 0, 1, 2, 9}, 5 * sizeof(int));
It's portable, but only to C99 implementations, which severely
undermines its practical value.
(I imagine that a trap representation might be generated after copying
some of the bytes of an int)

No, objects may always be safely manipulated as arrays of
character type, and this is the way that memcpy copies.
--
Here's a tip: null pointers don't have to be *dull* pointers!