Thread: Passing stucts by reference, malloc

Alright, I'm relatively new to C but not to programming. What I'm trying to do is create a struct, pass it to another function that will allocate memory for it and fill it.

Here's an example of what I'm doing, which creates exactly the same situation that I am dealing with

Code:

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

typedef struct _dog {
	char age;
	char weight;
} dog;

void makenewdog(dog * dogtomake) {
	dogtomake = (dog *) malloc(sizeof(dog));
	dogtomake->age = 1;
	dogtomake->weight = 5;
        /* this prints 5, which is correct */
	printf("Ok, I made you a dog. It's %d pounds.\n", dogtomake->weight);
}

int main (int argc, const char * argv[]) {
	dog doggie;
	doggie.weight = 0; doggie.age = 0;
	
	makenewdog(&doggie);
	
	/* weight should now be 5, but it returns 0. doggie's pointer isn't changed to
	   respect what it was given in makenewdog either */
	printf("I have a dog now, %d pounds.\n", doggie.weight);
	return 0;
}

Any pointers (no pun intended) in the right direction would be a great help, thanks!

You probably need to create a pointer to the dog struct in order for it to work.

Try changing to this:

Code:

int main (int argc, const char * argv[]) {
	dog * doggie;
	
	makenewdog(doggie);
	
	printf("I have a dog now, %d pounds.\n", doggie -> weight);
	return 0;
}

[EDIT]

Oh, and don't cast malloc - http://faq.cprogramming.com/cgi-bin/...&id=1043284351

If you wish to allocate the memory in the called function, you will have to pass a pointer to a pointer as demonstrated:

Code:

void makenewdog(dog ** dogtomake) {
        dog* tempDog; /* Don't need to use a temp but provides cleaner code. */

	tempDog = malloc(sizeof(dog));
	tempDog->age = 1;
	tempDog->weight = 5;
        /* this prints 5, which is correct */
	printf("Ok, I made you a dog. It's %d pounds.\n", tempDog->weight);

        *dogtomake = tempDog; /* Store the value of tempDog in the pointer provided by the calling function. */
}

int main (int argc, const char * argv[]) {
	dog * doggie; /* Do not use this until allocated in makenewdog. */

	makenewdog(&doggie); /* Pass the address of a dog pointer. */
	
	/* weight should now be 5, but it returns 0. doggie's pointer isn't changed to
	   respect what it was given in makenewdog either */
	printf("I have a dog now, %d pounds.\n", doggie->weight);
	return 0;
}

The other option is to return the dog pointer as the return value of the makenewdog function.

I would say that is often not a good idea to allocate the structure in the called function. The calling function can be responsible for allocation (either using malloc or on the stack) and then pass it to a function to complete its data if required.

EDIT: bivhitscar, remember that C always passes function arguments by value. In your example, the value passed to makenewdog is the uninitialised random value of the doggie variable.

[edit]
I was replying to #2, now 3

Thanks anonytmouse, that worked perfectly. I sort of had an idea that I would need to do something like that, but I've never worked with pointers to pointers before.

Thanks for the assistance, you've been a great help!

Ignore my solution, I was wrong. :P

But you still shouldn't cast malloc.

Try again bivhitscar

Code:

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

typedef struct _dog {
	char age;
	char weight;
} dog;

dog *makenewdog() {
	dog * dogtomake;
	dogtomake = malloc(sizeof(*dog));
	dogtomake->age = 1;
	dogtomake->weight = 5;
        /* this prints 5, which is correct */
	printf("Ok, I made you a dog. It's %d pounds.\n", dogtomake->weight);
	return dogtomake;
}

int main (int argc, const char * argv[]) {
	dog *doggie;
	
	doggie = makenewdog();
	
	/* weight should now be 5, but it returns 0. doggie's pointer isn't changed to
	   respect what it was given in makenewdog either */
	printf("I have a dog now, %d pounds.\n", doggie->weight);
	return 0;
}

Thanks salem; my thought processes are still developing.