Thread: Arrays of Structures in a Structure

Originally Posted by DTM

I haven't programed in C for a long time and I have decieded on a simple project to get myself back into it. However, it seems mot so simple!

I want to create a list of pupil details in a number of classes. So far i've got:

Code:

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

typedef struct {
	char * name;
	int age;
} pupil;

typedef struct {
	char * name;
	int num_pupils;
	struct pupil ** pupils;
} class;

int main ()
{
	class * class1;
	
	class1 = (class *)calloc(sizeof(class),1);

You don't need this cast. also, calloc() will stick all the bits to 0 which is not guaranteed to set the pointers to the null constant pointer (aka NULL). Finally, *alloc() can fail. You must test the returned value.

Code:

	class1->name = (char *)calloc(sizeof(char),128);

sizeof (char) is 1 by-definition.

Code:

	strcpy (class1->name, "Hello");
	class1->num_pupils = 20;
	printf ("Class %s has %d pupils\n", class1->name, class1->num_pupils);

	class1->pupils = (pupil*)calloc(sizeof(pupil), 20);

The cast is wrong. Just drop it. You want

Code:

	class1->pupils = malloc (sizeof *class1->pupils * 20); /* fixed */

Then, you have to set the 20 pointers to NULL, and to record the pupil's address... But you know that (I think).

Code:

}

Don't forget to free the memory when you don't use it any more.

This little project would be a good support to implement some smart ADTs (Abstract Data Types).

Originally Posted by DTM

Thanks for that,

I've tried to compile the code again but get the error message:

"invalid operands to binary * "

for the last line:

Code:

	class1->pupils = malloc (sizeof *class1->pupils) * 20;

Because it's darn wrong! I meant:

Code:

	class1->pupils = malloc (sizeof *class1->pupils * 20);

(Original fixed too)

You're trying to multiply malloc()'s return address by 20. This is bad. Move the *20 inside the malloc() function:
class1->pupils = malloc(sizeof(*class1->pupils)*20);

Originally Posted by itsme86

You're trying to multiply

Actually, I was trying. Code fixed.

Thanks

class1->pupil[0] is a pointer. As such, you just assign it the pointer directly:

Code:

pupil * pupil_1;

pupil_1 = malloc( sizeof *pupil_1 );
class1->pupil[0] = pupil_1;

Assuming you've allocated your class1->pupil "array".

Quzah.

Originally Posted by DTM

Maybe it's been too long since i've coded in C but the code for assigning the adress of the pupils should look like:

Code:

	pupil * pupil_1;

	pupil_1 = malloc (sizeof(pupil);

	class1->pupil[0] = &pupil_1;

or have i done something stupid?

I leave the comments on your own...

A pointer is a variable that holds an address. You just want to copy the value.

Code:

{
   pupil * pupil_1 = malloc (sizeof *pupil_1);

   if (pupil_1 != NULL)
   {
      class1->pupil[0] = pupil_1;
   }
}

Since you didn't post any code as to your latest error, and since you apparently didn't grasp the intent of my pervious post:

Code:

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

struct foo;
struct bar;

struct foo {
        struct bar**b;
};

struct bar
{
        int x;
};

int main( void )
{
        struct foo f, *p = &f;
        struct bar *b;
        int x;

        p->b = malloc( sizeof *b * 3 );
        for( x = 0; x < 3; x++ )
        {
                b = malloc( sizeof *b );
                p->b[x] = b;
        }

        for( x = 0; x < 3; x++ )
        {
                free( p->b[x] );
        }
        free( p->b );

        return 0;

}

You get the lazy version, because I didn't feel like mallocing and freeing foo.

Quzah.

Code:

p->b = malloc( sizeof *b * 3 );

I think you meant:

Code:

p->b = malloc( sizeof *p->b * 3 );

Code:

 struct foo;
 struct bar;

You don't need to declare these two here (just saving those typing fingers! )

Originally Posted by Hammer

Code:

p->b = malloc( sizeof *b * 3 );

I think you meant:

Code:

p->b = malloc( sizeof *p->b * 3 );

:doh:

Quzah.