How to duplicate a structure into a pointer with malloc function [C-Programming]

Question

I've tried to create a function that duplicate a structure into a pointer, but here is the issue, there is a char tab into the structure and I can't assign my original value to the new structure.

The function :

Planete *dupliquer(Planete *p){
  Planete *res;  
  res = (Planete*) malloc(sizeof(Planete));
  if(res == NULL){
    printf("Erreur d'allocation...\n");
    exit(1);
  }
  res->nomplanete = p->nomplanete;
  res->rayon = p->rayon;

  return res;
}

And here is the compilator error :

 error: incompatible types when assigning to type ‘char[20]’ from type ‘char *’
   res->nomplanete = p->nomplanete;
                   ^

Can you please help me, It will be very nice. Thanks for your support !

Thanks but I have to use the malloc function, do you have any solution for my error ? — oktomus, Mar 02 '15 at 15:36
@tux3 has already given a simple solution, but for future reference note that your problem is caused by trying to use `=` to assign char * strings (you would need to use strcpy or similar for this). Also note that [you should not cast the result of malloc in C](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). — Paul R, Mar 02 '15 at 15:36
Don't cast the return value of `malloc()` it's not necessary. — Iharob Al Asimi, Mar 02 '15 at 15:37
I create a Planete, the I duplicate it into a pointer I created and allocate into memory — oktomus, Mar 02 '15 at 15:37
I don't really understand the cast behind the malloc, this is my teacher instruction ^^ — oktomus, Mar 02 '15 at 15:45
@KevinMasson Your teacher is bad, tell them to go [read this](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). — Lundin, Mar 02 '15 at 15:45
Well, this is a major problem in the whole software industry: way too many crappy teachers and crappy books. — Lundin, Mar 02 '15 at 15:49
I shown it to my teacher and he didn't understand why it works x) Thank you :) — oktomus, Mar 02 '15 at 15:51
Your teacher is probably confused between C and C++ - in C you should not use a cast with malloc, in C++ you have to (but then you shouldn't be using malloc in C++ anyway). — Paul R, Mar 02 '15 at 22:33

Sergey Kalinichenko · Accepted Answer · 2015-03-02T15:43:15.697

2

It looks like you do not need a separate malloc for nomplanete, because it is an array. In situations like that you should use strcpy, like this:

strcpy(res->nomplanete, p->nomplanete);

If all members of Planete are primitives or arrays, you could simply memcpy the entire thing, like this:

res = malloc(sizeof(Planete)); // No need to cast
if(res == NULL){
    printf("Erreur d'allocation...\n");
    exit(1);
}
memcpy(res, p, sizeof(Planete));

If nomplanete were a pointer, you would have to do a separate malloc or use strdup:

res->nomplanete = malloc(1+strlen(p->nomplanete));
strcpy(res->nomplanete, p->nomplanete);

edited Mar 02 '15 at 15:43

answered Mar 02 '15 at 15:35

Sergey Kalinichenko

714,442
84
1,110
1,523

If "nomplanete" was an array as member of the struct, then memcpy() on the whole struct would have worked just fine. – Lundin Mar 02 '15 at 15:40

score 0 · Answer 2 · answered Mar 02 '15 at 15:44

0

It appears that "nomplanete" is a character array member of the "Planete" struct. In that case, simply write the function as:

Planete* dupliquer (const Planete* p)
{
  Planete* res;  
  res = malloc(sizeof(Planete));
  if(res == NULL){
    printf("Erreur d'allocation...\n");
    exit(1);
  }
  memcpy(res, p, sizeof(Planete));

  return res;
}

answered Mar 02 '15 at 15:44

Lundin

195,001
40
254
396

Finally, I've used strcpy beceause we had to do it without memcpy ^ ^ Thanks ! – oktomus Mar 02 '15 at 15:54

How to duplicate a structure into a pointer with malloc function [C-Programming]

2 Answers2