pointer and struct

**toshog** · 10-11-2004

hi all

i'm learning the use of pointers within struct definitions. currently following the instructions here:
http://www.le.ac.uk/cc/tutorials/c/ccccstrc.html

my code below produces:
syntax error before "struct" on line 15 (maked)

can somebody please explain why?

thanks a lot..

Code:

main() {


        struct p {

        int data;
        int voo, loo, soo;
        struct p *moo;

        }*koo;

>>15>>  koo=(*struct p) malloc(sizeof(*koo));

        koo->data=333;

}

**misplaced** · 10-11-2004

TRY koo= (struct p *) malloc(sizeof(p));

i may be wrong

**toshog** · 10-11-2004

Originally Posted by misplaced

TRY koo= (struct p *) malloc(sizeof(p));

i may be wrong

thanks... that worked. i notoced that the last update on the page i'm following instructions from is from 3 and a half years ago. has something in the annotation of the type casting changed or it's just a typing error on that page? how come the * is after the struct name in the casting?

thanks...

**Micko** · 10-11-2004

I suggest changing tutorial.
I recommend:
http://faq.cprogramming.com/cgi-bin/...&id=1073086407

**Salem** · 10-11-2004

That is one sorry looking tutorial you have there.
So many faults, and I've only skimmed one page of it.

**Brain Cell** · 10-11-2004

casting 'malloc()' is not needed in a C code. It works fine for me in Dev C++ under C projects.

**toshog** · 10-12-2004

thanks.... so far so good...

another question. here is a piece of code:

Code:

main() {


typedef struct p {

        int data;
        int voo, soo;
        struct p *moo;

        }koo;

        koo *loo;
        loo->data=333;
        loo->voo=loo->soo=999;

       void print_it(koo *some_pointer) {
                printf("the thing is %d\n",some_pointer->soo);
        }

        print_it(loo);
//      print_it(&loo);
   //     printf("the int is %d\n",loo->soo);
}

here loo is pointing to the address of a variable of type koo, right?
so how do i access the data variable of the *moo pointer in loo?
does this make sense?

also if i pass the loo to print_it by refernce - like print_it(&loo) - gcc says:
warning: passing arg 1 of `print_it' from incompatible pointer type and it prinst a wrong number?
according to the suggested tutorial that's how pointers to structures are passed to functions...

i'm trying to figure out all about the struct weired stuff, so please bare with me...

thanks....

**Salem** · 10-12-2004

Well since you have a nested function, I'm surprised that it compiles
Also, since your pointers are uninitialised, I'm surprised that it runs.

This also means that the answers you are printing could be bogus for reasons nobody can figure out.

Start by posting actual code you compiled, not random snippets of stuff you think is important arranged in some order bearing no resemblance to the code you compiled.

**toshog** · 10-12-2004

this was actually the whole file without the include statement. here it is again:

Code:

#include <stdio.h>

main() {


typedef struct p {

        int data;
        int voo, soo;
        struct p *moo;

        }koo;

        koo *loo;
        loo->data=333;
        loo->voo=loo->soo=999;

        void print_it(koo *some_pointer) {
                printf("the thing is %d\n",some_pointer->soo);
        }


        print_it(loo);

 //     printf("the int is %d\n",loo->soo);
}

i understand what you are saying about the nested function...
this code actually compiles and runs and it works fine...

isn't the line "koo *loo" initiate the loo pointer? declaring pointer to loo of type koo?

thanks....

**quzah** · 10-12-2004

Sure, it declares a pointer. But what is your pointer pointing at? Try allocating some memory for the actual structure instance before you try and use your pointer. And no, it doesn't "work[s] fine", it "lucks out".

Quzah.

**Salem** · 10-12-2004

Heh - so many problems

Code:

gcc -W -Wall -ansi -pedantic -O2  hello.c
hello.c:3: warning: return type defaults to `int'
hello.c: In function `main':
hello.c:18: warning: ISO C forbids nested functions
hello.c:20: warning: ISO C89 forbids mixed declarations and code
hello.c:26: warning: control reaches end of non-void function
hello.c:14: warning: `loo' might be used uninitialized in this function

Start beefing up your compiler options - they can tell you quite a lot about the problems you have.

Try this

Code:

#include <stdio.h>

typedef struct p {
    int data;
    int voo, soo;
    struct p *moo;
} koo;

void print_it(koo * some_pointer)
{
    printf("the thing is %d\n", some_pointer->soo);
}

int main()
{
    koo bar;
    koo *loo = &bar;
    loo->data = 333;
    loo->voo = loo->soo = 999;
    print_it(loo);
    return 0;
}

**toshog** · 10-13-2004

thank you. i appreciate the corrections. one last thing:
why are these declarations needed:

Code:

    koo bar;
    koo *loo = &bar;

sorry about the stupid question... just trying to clear the fog... thank you.

**Micko** · 10-13-2004

Originally Posted by toshog

thank you. i appreciate the corrections. one last thing:
why are these declarations needed:

Code:

    koo bar;
    koo *loo = &bar;

sorry about the stupid question... just trying to clear the fog... thank you.

Because in first line you declare a variable of type koo and in the second line you decalre a pointer variable named loo. Because loo is a pointer variable it needs to point to something which means that the value of loo should be address of variable bar.
So the second line initialize pointer to type koo with addressof variable bar.

**toshog** · 10-13-2004

Originally Posted by Micko

Because in first line you declare a variable of type koo and in the second line you decalre a pointer variable named loo. Because loo is a pointer variable it needs to point to something which means that the value of loo should be address of variable bar.
So the second line initialize pointer to type koo with addressof variable bar.

makes sense. i think i understand now... thank you.

**misplaced** · 10-13-2004

lol........i love it...forget the inline keyword.....just go ahead and put your functions inline

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