dynamic declaration of struct array

[Tim]

Why won't the declaration of a struct array work if I do it like this:

1 typedef struct
2 {
3 int pens;
4 int pencils;
5 } Stationers;
6

.........

7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

[Noah Roberts]

Tim wrote:
Stationers stats[num];

12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

Stationers *stats = (Stationers*)ma lloc(num * sizeof(Statione rs));

NR

[Ivan Vecerina]

"Tim" <hi**********@y ahoo.co.uk> wrote in message
news:ad******** *************** ***@posting.goo gle.com...

Why won't the declaration of a struct array work if I do it like this: .... 7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

This code is actually valid if your compiler implements the 1999
C standard. Prior to that, array variables are indeed required
to be of a size known at compile time.

The backwards-compatible way to do what you want is to use:
Stationers* stats = (Stationers*)ma lloc(num*sizeof (Stationers));
... use stats as if it were an array of size num ...
free(stats); // don't forget this prior to exit !
Hope this helps,
Ivan
--
http://ivan.vecerina.com

[Dave Vandervies]

In article <ad************ **************@ posting.google. com>,
Tim <hi**********@y ahoo.co.uk> wrote:

Why won't the declaration of a struct array work if I do it like this:
[snip struct definition]
7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

Uhmm... what language are you trying to write this code in?

If you're using C99, what you have here should work, since arrays are
allowed to have lengths that aren't known until run-time.

If you're using C90, you're not allowed to declare a new variable (like
the array) after non-declaration stuff (like assigning a value to num),
so once you fill in the blanks your code will still be broken no matter
how you declare the array. Instead, you'd need to declare a pointer at
the beginning of the block and, once you've decided how many elements
you want, use malloc to allocate memory for it:
--------
#include <stdlib.h>

/*Define and typedef the Stationers struct here*/

int main(void)
{
int num;
Stationers *stats;

/*Decide on a value for num here*/

stats=malloc(nu m * sizeof *stats);

/*I assume you'll want to do something with stats here*/

free(stats);
}
--------

If you're using C++ (which also allows declarations after non-
declarations), go to comp.lang.c++ and ask about (or, better, open a
textbook and read about) new[] and std::vector.
dave

--
Dave Vandervies dj******@csclub .uwaterloo.ca
So "Patterns" are really just another name for "Object-oriented programming",
which is really just another name for "Structured programming", which is really
just big words for "good hacking". --Richard Bos in comp.lang.c

[Irrwahn Grausewitz]

hi**********@ya hoo.co.uk (Tim) wrote:

Why won't the declaration of a struct array work if I do it like this:

1 typedef struct
2 {
3 int pens;
4 int pencils;
5 } Stationers;
6

.........

7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

Yes, you have to define a pointer and dynamically allocate memory for
it to point to, look:

#include <stdio.h>
#include <stdlib.h>

typedef
struct
{
int pens;
int pencils;
}
Stationers;

int main( void )
{
int num;
Stationers *stats;

/* calculate value for num, e.g: */
num = 42;

stats = malloc( num * sizeof *stats );
if ( stats == NULL )
{
fprintf( stderr, "Memory allocation error." );
return EXIT_FAILURE;
}

/* do something with stats, e.g: */
stats[ 7 ].pens = 6;
stats[ 7 ].pencils = 9;

return EXIT_SUCCESS;
}

Regards
--
Irrwahn
(ir*******@free net.de)

[Irrwahn Grausewitz]

dj******@csclub .uwaterloo.ca (Dave Vandervies) wrote:

Tim <hi**********@y ahoo.co.uk> wrote: <SNIP>

7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

Uhmm... what language are you trying to write this code in?

If you're using C99, what you have here should work, since arrays are
allowed to have lengths that aren't known until run-time.

If you're using C90, you're not allowed to declare a new variable (like
the array) after non-declaration stuff (like assigning a value to num),
so once you fill in the blanks your code will still be broken no matter
how you declare the array. Instead, you'd need to declare a pointer at
the beginning of the block and, once you've decided how many elements
you want, use malloc to allocate memory for it:
--------
#include <stdlib.h>

/*Define and typedef the Stationers struct here*/

int main(void)
{
int num;
Stationers *stats;

/*Decide on a value for num here*/

stats=malloc(nu m * sizeof *stats);

/* check malloc() return value here! */
/*I assume you'll want to do something with stats here*/

free(stats); return 0;}

<SNIP>
--
Irrwahn
(ir*******@free net.de)

[Irrwahn Grausewitz]

Noah Roberts <nr******@donte mailme.com> wrote:

Tim wrote:

[ some code restored ... ]

7 int main(void)
8 {
9 int num;
...........num defined in code........
10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

Stationers *stats = (Stationers*)ma lloc(num * sizeof(Statione rs));

Better:

Stationers *stats = malloc(num * sizeof *stats);

(The cast to is not only unnecessary, it may hide the failure of not
including stdlib.h; sizeof *stats has the advantage that one does not
have to change the malloc expression if the type of stats ever changes.)

But still this won't work in pre-C99 (in this special case): you define
a variable in another place than the beginning of a block. Or, if you
intended to put above line at the top of main(), you don't know the
value for num, as it has to be calculated yet.

So, there's no way around splitting the line up into:

/* goes to top of main(): */
Stationers *stats;

/* after num has been calculated: */
stats = malloc(num * sizeof *stats);

Regards
--
Irrwahn
(ir*******@free net.de)

[Irrwahn Grausewitz]

"Ivan Vecerina" <ivecATmyrealbo xDOTcom> wrote:

"Tim" <hi**********@y ahoo.co.uk> wrote in message
news:ad******* *************** ****@posting.go ogle.com...

Why won't the declaration of a struct array work if I do it like this:...

7 int main(void)
8 {
9 int num;

...........num defined in code........

10
11 Stationers stats[num];
12 }

The error message says "constant expression required" and highlights
line 11. I want to declare it dynamically during runtime, could
someone explain how to do this please.

This code is actually valid if your compiler implements the 1999
C standard. Prior to that, array variables are indeed required
to be of a size known at compile time.

The backwards-compatible way to do what you want is to use:
Stationers* stats = (Stationers*)ma lloc(num*sizeof (Stationers));

Better:

Stationers *stats = malloc(num * sizeof *stats);

(The cast is not only unnecessary, it may hide the failure of not
including stdlib.h; sizeof *stats has the advantage that one does not
have to change the malloc expression if the type of stats ever changes.)

But still this won't work in pre-C99 (in this special case): you define
a variable in another place than the beginning of a block. Or, if you
intended to put above line at the top of main(), you don't know the
value for num, as it has to be calculated yet.

So, there's no way around splitting the line up into:

/* goes to top of main(): */
Stationers *stats;

/* after num has been calculated: */
stats = malloc(num * sizeof *stats);
... use stats as if it were an array of size num ...
free(stats); // don't forget this prior to exit !

Regards
--
Irrwahn
(ir*******@free net.de)

[Ivan Vecerina]

"Irrwahn Grausewitz" <ir*******@free net.de> wrote in message
news:la******** *************** *********@4ax.c om...

int main( void )
{
int num;
Stationers *stats;

/* calculate value for num, e.g: */
num = 42;

stats = malloc( num * sizeof *stats );
if ( stats == NULL )
{
fprintf( stderr, "Memory allocation error." );
return EXIT_FAILURE;
}

/* do something with stats, e.g: */
stats[ 7 ].pens = 6;
stats[ 7 ].pencils = 9;
add:
free( stats );
return EXIT_SUCCESS;
}

Yes, the memory will be reclaimed at program exit on
any decent OS, but it is a good habit to always free
the memory that you allocate. (e.g. in case the code
gets moved into a loop or another function...).

Cheers,
Ivan
--
http://ivan.vecerina.com

[Irrwahn Grausewitz]

"Ivan Vecerina" <ivecATmyrealbo xDOTcom> wrote:

"Irrwahn Grausewitz" <ir*******@free net.de> wrote in message
news:la******* *************** **********@4ax. com...

int main( void )
{
int num;
Stationers *stats;

/* calculate value for num, e.g: */
num = 42;

stats = malloc( num * sizeof *stats );
if ( stats == NULL )
{
fprintf( stderr, "Memory allocation error." );
return EXIT_FAILURE;
}

/* do something with stats, e.g: */
stats[ 7 ].pens = 6;
stats[ 7 ].pencils = 9;
add:
free( stats );

return EXIT_SUCCESS;
}

Yes, the memory will be reclaimed at program exit on
any decent OS, but it is a good habit to always free
the memory that you allocate. (e.g. in case the code
gets moved into a loop or another function...).

Good point.

Cheers,
Ivan

Regards
--
Irrwahn
(ir*******@free net.de)