double pointer to structure

[ravi.xolve]

#include <stdio.h>
#include <stdlib.h>
struct a
{
        int a;
        char c;
};
main()
{
        struct a *p, q;
        void func(struct a **);
        printf("Previous address p: %u",p);
        func(&p);
        printf("\nNew address p: %u",p);
        printf("\nValues stored in p: %d %c\n",p->a,p->c);
}

void func(struct a **ptr)
{
        *ptr=(struct a *)malloc(sizeof(struct a));
        printf("\n*********%u",*ptr);
        *ptr->a=100;
        *ptr->c='f';
        /*try reducing and addind the *s */
}

why the above code gives the error and i cant help it while the code is correct logically. pls help soooooooon!

[gorda001]

Just a quick note before HC posts his large explanation...

It really is a bad idea to write code like:

struct a *p, q;

Because it's really unclear. Instead write:

struct a *p;
struct a q;

[Mythic]

Just a quick note before HC posts his large explanation...

Yeah! I'm gonna stop replying to c or c++ related posts, his answers always go into so much detail that mine just look stupid :-P

[gorda001]

It's funny that we are having this discussion still while he is writing :)

[hot_cakes]

Within func(), your printf statement has a conversion string that specifies an unsigned int is going to be given, but the argument you give is *ptr which is of type struct a* and not unsigned int.

You make similar printf calls elsewhere. To print the address represented by a pointer use %p, not %u. You'll need to cast your second arguments to printf to void*, too.

Where you have for example, *ptr->a=100, you should instead have (*ptr)->a=100 because -> takes precedence over * when used as a dereferencing operator.

You shouldn't cast the return value of malloc. Some older books will tell you to do so, but it's best avoided because it might hide errors. malloc (and calloc and realloc) return void* so there's no need to a cast.

Don't forget to free what you have malloc-ed, though you probably know that.

'a' probably isn't a very good name for a struct, especially if that struct has a member named 'a'!

A more idiomatic way of doing what you're doing is:

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

struct my_type
{
    int x;
};

struct my_type *create_my_type(int value)
{
    struct my_type *ret = malloc(sizeof *ret);
    if (ret) /* don't forget that malloc can fail!! */
        ret->x = value;

    return ret;
}

int main(void)
{
    struct my_type *p = create_my_type(5);
    if (p)
        printf("p is %p, p->x is %d\n", (void *)p, p->x);
    else
        printf("Failed to create a my_type");

    free(p);

    return 0; 
}

Edd

[hot_cakes]

It's funny that we are having this discussion still while he is writing :)

Grrr ;)

Edd

[Mythic]

Grrr ;)

Edd

I bet IRL you're just like JP from that film Grandma's Boy!! :)

That's not an insult, incase you've not seen the film!