Structure memory allocation

[Trying_Harder]

Consider the following declaration,

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

typedef struct foo {
char name[30];
int age;
}Foo;

typedef struct bar {
char * name;
int age;
}Bar;

typedef struct baz {
char name[30];
int *age;
}Baz;

int main()
{
Foo f1,f2;
Bar b1,b2;
Baz c1,c2;

/* Populating structure */
strcpy(f1.name,"JACK");
f1.age=10;

/* CASE 1 - Works fine */
f2=f1;
printf("\n F2 Members, Name %s , Age %d ", f2.name, f2.age);
printf("\n F1 Pointer %p , F2 Pointer %p ", f1, f2);
/* CASE 2 - Hmmmm.. I don't have an explanation for this */
b1.name = malloc(sizeof(char)*10);
strcpy(b1.name,"LACK");
b1.age=20;

b2=b1; /*Assignment??*/
printf("\n B2 Members, Name %s , Age %d ", b2.name, b2.age);
printf("\n B1 Pointer %p , B2 Pointer %p ", b1, b2);
/* This will print the same address out. */
printf("\n B1 Name Pointer %p , B2 Name Pointer %p ", b1.name, \
b2.name);
/* CASE 3 - As expected, does not work */
c1.age = malloc(sizeof(int));
strcpy(c1.name,"MACK");
*(c1.age)=30;

c2=c1; /*Assignment??*/
printf("\n C2 Members, Name %s , Age %d ", c2.name, c2.age);
printf("\n C1 Pointer %p , C2 Pointer %p ", c1, c2);

/* But this will print the same address out, how come? */
printf("\n C1 Name Pointer %p , C2 Name Pointer %p ", c1.age ,\
c2.age );

exit(EXIT_SUCCESS);
}
Let me begin with apologies for posting so much of code.

Clarification 1 : Is copying structures by simply assigning them, valid
(legal) portable, pedantic ?
<My Opinion> : No. If someone argues, please reason.

Clarification 2 : Why is case 2 working ?

Clarification 3 : In case 3, you will notice the address copied exactly.
Does copying address mean pointing to the same
location?

[Morris Dovey]

Trying_Harder wrote:

Clarification 1 : Is copying structures by simply assigning them, valid
(legal) portable, pedantic ?
<My Opinion> : No. If someone argues, please reason.
It's legal and portable (see standard)
Clarification 2 : Why is case 2 working ?
Why would you not expect it to work?
Clarification 3 : In case 3, you will notice the address copied exactly.
Does copying address mean pointing to the same
location?

It does, but you'd do better with:

printf("\n C2 Members, Name %s , Age %d ", c2.name, *c2.age);

since c2.age is a pointer (as you remembered in your next statement.

--
Morris Dovey
West Des Moines, Iowa USA
C links at http://www.iedu.com/c

[Al Bowers]

Trying_Harder wrote:

Consider the following declaration,

#include <stdio.h>
#include <stdlib.h>
#include <string.h> /* for function strcpy */

typedef struct foo {
char name[30];
int age;
}Foo;

typedef struct bar {
char * name;
int age;
}Bar;

typedef struct baz {
char name[30];
int *age;
}Baz;

int main()
{
Foo f1,f2;
Bar b1,b2;
Baz c1,c2;

/* Populating structure */
strcpy(f1.name,"JACK");
f1.age=10;

/* CASE 1 - Works fine */
f2=f1;
printf("\n F2 Members, Name %s , Age %d ", f2.name, f2.age);
printf("\n F1 Pointer %p , F2 Pointer %p ", f1, f2);

%p specifies a void* argument. f1 and f2 need to be pointers that
should be cast to void*.
printf("\n F1 Pointer %p, F2 Pointer %p ", (void*)&f1, (void*)&f2);
/* CASE 2 - Hmmmm.. I don't have an explanation for this */
b1.name = malloc(sizeof(char)*10);
strcpy(b1.name,"LACK");
b1.age=20;

b2=b1; /*Assignment??*/
printf("\n B2 Members, Name %s , Age %d ", b2.name, b2.age);
printf("\n B1 Pointer %p , B2 Pointer %p ", b1, b2);
/* This will print the same address out. */
printf("\n B1 Name Pointer %p , B2 Name Pointer %p ", b1.name, \
b2.name);

/* CASE 3 - As expected, does not work */
c1.age = malloc(sizeof(int));
strcpy(c1.name,"MACK");
*(c1.age)=30;

c2=c1; /*Assignment??*/
printf("\n C2 Members, Name %s , Age %d ", c2.name, c2.age);
*c2.age
printf("\n C1 Pointer %p , C2 Pointer %p ", c1, c2);

/* But this will print the same address out, how come? */
printf("\n C1 Name Pointer %p , C2 Name Pointer %p ", c1.age ,\
c2.age );
Perhaps you are being confused by the above printf statement. You
are printing "Name Pointer" but your arguments refer to the age members.
Shouldn't these be:

printf("C1 Name Pointer %p , C2 Name Pointer %p\n",
(void*) c1.name ,(void*)c2.name);
printf("C1 Age Pointer %p , C2 Age Pointer %p\n", (void*)c1.age ,
(void*)c2.age );
exit(EXIT_SUCCESS);
}


Clarification 3 : In case 3, you will notice the address copied exactly.
Does copying address mean pointing to the same
location?

Try the corrected code and see if you still need clarification.

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

typedef struct foo {
char name[30];
int age;
}Foo;

typedef struct bar {
char * name;
int age;
}Bar;

typedef struct baz {
char name[30];
int *age;
}Baz;

int main(void)
{
Foo f1,f2;
Bar b1,b2;
Baz c1,c2;

strcpy(f1.name,"JACK");
f1.age=10;
f2=f1;
printf("F2 Members, Name %s , Age %d\n", f2.name, f2.age);
printf("F1 Pointer %p , F2 Pointer %p\n", (void*)&f1, (void*)&f2);
b1.name = malloc(sizeof(char)*10);
strcpy(b1.name,"LACK");
b1.age=20;
b2=b1; /*Assignment??*/
printf("B2 Members, Name %s , Age %d\n", b2.name, b2.age);
printf("B1 Pointer %p , B2 Pointer %p\n", (void*)&b1, (void*)&b2);
/* This will print the same address out. */
printf("B1 Name Pointer %p , B2 Name Pointer %p\n",
(void*)b1.name, (void*)b2.name);
c1.age = malloc(sizeof(int));
strcpy(c1.name,"MACK");
*(c1.age)=30;
c2=c1; /*Assignment??*/
printf("C2 Members, Name %s , Age %d\n", c2.name, *c2.age);
printf("C1 Pointer %p , C2 Pointer %p\n", (void*)&c1, (void*)&c2);
printf("C1 Name Pointer %p , C2 Name Pointer %p\n",
(void*) c1.name ,(void*)c2.name);
printf("C1 Age Pointer %p , C2 Age Pointer %p\n", (void*)c1.age ,
(void*)c2.age );
return 0;
}

--
Al Bowers
Tampa, Fl USA
mailto: xa*@abowers.combase.com (remove the x)
http://www.geocities.com/abowers822/

[Jake Roersma]

On Mon, 15 Sep 2003 22:30:44 -0700, Trying_Harder wrote:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct baz {
char name[30];
int *age;
}Baz;

int main()
{
Foo f1,f2;
Bar b1,b2;
Baz c1,c2;

/* Populating structure */
strcpy(f1.name,"JACK");
f1.age=10;

/* CASE 1 - Works fine */
printf("\n F1 Pointer %p , F2 Pointer %p ", f1, f2);
This may work but is not right. Try:

printf("\n F1 Pointer %p , F2 Pointer %p ", &f1, &f2);

printf("\n B1 Pointer %p , B2 Pointer %p ", b1, b2);
And again....

printf("\n B1 Pointer %p , B2 Pointer %p ", &b1, &b2);

printf("\n C2 Members, Name %s , Age %d ", c2.name, c2.age);
You'll need to dereference c2.age to get its value:

printf("\n C2 Members, Name %s , Age %d ", c2.name, *c2.age);
printf("\n C1 Pointer %p , C2 Pointer %p ", c1, c2);
I hate to be repetitive.. but...

printf("\n C1 Pointer %p , C2 Pointer %p ", &c1, &c2);

/* But this will print the same address out, how come? */
printf("\n C1 Name Pointer %p , C2 Name Pointer %p ", c1.age ,\
c2.age );

free(b1.name);
free(c1.age);
exit(EXIT_SUCCESS);
}
Let me begin with apologies for posting so much of code.

Clarification 1 : Is copying structures by simply assigning them, valid
(legal) portable, pedantic ?
legal and portable.

Clarification 2 : Why is case 2 working ?
Other than the fact you are not freeing the memory you've allocated I
don't see any reason as to why it shouldn't work.
Clarification 3 : In case 3, you will notice the address copied exactly.
Does copying address mean pointing to the same
location?

Yep.

- Jake

[Irrwahn Grausewitz]

Jake Roersma <ja***@copiosus.net> wrote:

On Mon, 15 Sep 2003 22:30:44 -0700, Trying_Harder wrote: <SNIP>

printf("\n F1 Pointer %p , F2 Pointer %p ", f1, f2);

This may work but is not right. Try:

printf("\n F1 Pointer %p , F2 Pointer %p ", &f1, &f2);

The printf %p conversion specification requires a void pointer argument;
so make it:

printf("\n F1 Pointer %p , F2 Pointer %p ", (void *)&f1, (void *)&f2);

printf("\n B1 Pointer %p , B2 Pointer %p ", b1, b2);

And again....

printf("\n B1 Pointer %p , B2 Pointer %p ", &b1, &b2);

.... and again ...

printf("\n B1 Pointer %p , B2 Pointer %p ", (void *)&b1, (void *)&b2);

printf("\n C2 Members, Name %s , Age %d ", c2.name, c2.age);

You'll need to dereference c2.age to get its value:

printf("\n C2 Members, Name %s , Age %d ", c2.name, *c2.age);

printf("\n C1 Pointer %p , C2 Pointer %p ", c1, c2);

I hate to be repetitive.. but...

printf("\n C1 Pointer %p , C2 Pointer %p ", &c1, &c2);

.... me too, but ... ;)

printf("\n C1 Pointer %p , C2 Pointer %p ", (void *)&c1, (void *)&c2);

<SNIP>
Note that the only useful application for printing out pointer values I
know of is for debugging purposes.

Regards

Irrwahn
--
What does this red button do?