Pointer in struct?

[LeonLanford]

Excuse me, I want to ask please. Sorry if my english is bad.

I have a problem, my lecturer wanted me to make a program which consists only pointer and malloc.

I have a problem in making variable in struct to pointer.

I mean like this

struct xampl {
  int x;
};

what my lecturer want is

struct xampl {
  int *x;
};

He said that i can use typedef, but even I tried so many times. I still failed.

Is anyone know how to solve this problem?

[KONI]

Well, what's the difference between this one and a normal pointer ? NOTHING

Just initialize a struct xampl and then do what you always do:

    struct xampl myStruct;
    myStruct.x = malloc (sizeof(int));
    *(myStruct.x) = 123;

[LeonLanford]

Sorry my computer was broken down so I cannot reply.

Thanks KONI! it's working

but I still confuse what is the difference between

*(myStruct.x) = 123;

and

*(myStruct).x = 123;

because both of it worked

[quzah]

'myStruct' isn't a pointer, so you don't dereference it. You dereference the pointer it contains.


Quzah.

[MacGyver]

Edited out until I get some coffee or something....

[AnotherEdit]

OK, another try. Don't know why I didn't realize that . has higher precedence than *. Obviously, that's why we have ->.

Overall, this means that both expressions in this case mean the same thing. Both of them take myStruct, then grab its member x (due to the .), dereference it (due to the *), and use that as the lvalue for the assignment.

For whatever reason I got it backwards. I suppose lack of sleep does that to people.

[/AnotherEdit]

[LeonLanford]

I'm getting more confused now..

so what is the right one?

both of it worked, I've tested it

[ZuK]

you didn't tell how you declared myStruct.
true with a cast you can make both "work" but only one is right.
Show the full code.
Kurt

[LeonLanford]

The code is like this

#include <stdio.h>

struct data
{
   int *bil;
} ;

int main()
{
   struct data *tdata;
   
    tdata = (struct data *) malloc (sizeof(struct data *));

    tdata.bil = (int *) malloc (sizeof(int));

    *(tdata).bil = 123;

    printf("%d", *(tdata).bil);    

system("pause");
return 0;
}

[MacGyver]

That code doesn't compile for me.

1) Missing stdlib.h.
2) Improper assignments with pointer types and all that stuff.

The notations we are discussing are valid provided that the struct variable you're dealing with is not a pointer. This means tdata should be of type struct data, not a pointer to struct data.

[LeonLanford]

I'm using devc++ 4.9.9.2

I usually don't put stlib because it does run in devc.

My lecturer said that it's pointer to struct, I'm getting more confused now..

[ZuK]

in that case

tdata->bil = (int *) malloc (sizeof(int));
*((*tdata).bil) = 123;

or

(*tdata).bil = (int *) malloc (sizeof(int));
*(tdata->bil) = 123;

would be right.
There shouldn't be a cast of malloc' s return value.
Without that the compiler would have warned you about the mistakes.
Kurt

[MacGyver]

Altering your code slightly.... This will compile:

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

struct data
{
	int *bil;
} ;

int main()
{
	struct data  *tdata;
	
	tdata = (struct data *)malloc(sizeof(*tdata));
	tdata->bil = (int *)malloc(sizeof(int));
	
	*tdata->bil = 123;
	
	printf("%d\n", *tdata->bil);
	
	system("pause");
	return 0;
}

[ZuK]

I usually don't put stlib because it does run in devc.

If you don't include stdlib.h there is no malloc defined.
That "works" only on a platform where pointers and int's have the same size but the compiler will give you wrong diagnostics.
Kurt

[LeonLanford]

hmm..

zuk says

*(tdata->bil) = 123;

and mac says

*tdata->bil = 123;

so both is right? sorry i cannot test the code because i'm now in internet cafe

[ZuK]

Yes both is right. Check out operator precedence.
Kurt