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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?

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

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

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: (remove the x)
http://www.geocities.com/abowers822/

Jake Roersma <> 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?