Values in array of structs turning into garbage values

Question

I have an issue that's really confusing me... Below I am calling an initialize function:

void Initialize (List *L) {
    char* initialize = "initialize";
    int i;

    for (i=0; i<MAXLISTSIZE; i++) {
       strncpy(L->items[i].name,initialize,MAXNAMESIZE);
       L->items[i].name[MAXNAMESIZE - 1] = '\0';
       L->items[i].grade = 0;
       printf("L->items[i].name = %s\n", L->items[i].name);
       printf("L->items[i].grade = %d\n", L->items[i].grade);
    }    
    L->count = 0;
}

And it seems to work, I print the values in the loop and it's fine. If I also print inside an identical loop in main to double check it works as well but If I just print the values in main after the initialize function (no print statements in Initialize) I get complete garbage.

It seems the memory I'm storing my values in isn't staying consistent and I can't figure out why.

Do I need to malloc memory for the structs? Since I don't need a variable amount of storage I thought it was not necessary... I am unsure of how to go about that.

My Structs:

typedef Student Item;
#define MAXLISTSIZE 4
typedef struct {
Item items[MAXLISTSIZE];
int count;
} List;

#define MAXNAMESIZE 20
typedef struct {
char name[MAXNAMESIZE];
int grade;   
} Student;

I am simply calling Initialize from main:

int main () {
    List *newList;

    /*call initialize function*/
    newList = callInitialize();

return 0;
}

callInitialize:

List *callInitialize () {
List *L;

List studentList;
L = &studentList;

Initialize(L); 

return L;
}

What's callInitialize and what does it have to do with Initialize??? Where and how do you call Initialize? — AnT stands with Russia
– AnT stands with Russia, Commented Oct 1, 2011 at 23:20
1) Always compile with all warnings. 2) Use valgrind or a similar memory debugger. Don't post until you've done both of those and come out clean, or do post the specific error or warnings that you get. The point is that if you do any amount of debugging work yourself, then you either discover the problem, or you'll be able to ask a specific question rather than dump your entire code and say "this doesn't work". — Kerrek SB
– Kerrek SB, Commented Oct 1, 2011 at 23:23
@Kerrek-SB I'm sorry I was missing a function and added that. I realize I can't expect to post my code and say fix it, this is not my whole code but everything I thought that was important to my question, I don't get any compiler warnings. I have spent a long time trying to get it working myself but haven't been able to come up with a solution. I am not a very experienced programmer and have never used a memory debugger but I will look into that. Sorry again, and thank you for your time. — cdn88
– cdn88, Commented Oct 1, 2011 at 23:57
@cdn88: No worries! What's your compiler? I'm not sure if you'd get a warning, since this is somewhat hard to detect statically. However, whenever you see any sort of "garbled output", you should definitely check out a memory debugger. On Linux, valgrind is extremely easy to use and exceedingly helpful. (Compile with debug symbols for best results.) I'm sure others can suggest memory debuggers for Windows and MacOS. — Kerrek SB
– Kerrek SB, Commented Oct 2, 2011 at 0:00
@Kerrek-SB: I am using gcc with these (-Wall -ansi -pedantic -g) compiler arguments. I am definitely going to check out valgrind! I wish I would have learned about that earlier. Thank you! — cdn88
– cdn88, Commented Oct 2, 2011 at 0:06

Kerrek SB · Accepted Answer · 2011-10-02 00:05:11Z

5

Now that you posted the function that causes the actual problem, we see what's wrong: You are returning the address of a local variable that goes out of scope! This is not valid.

List * foo()
{
  List x;     // <--- x comes to life
  return &x;  // <--- x dies here...
}

int main()
{
  List * p = foo();  // ... but its address is used here!
  p->name ...        // dereferencing an invalid address!!
}

Your situation calls for dynamic (or "manual") allocation, which means memory allocation whose lifetime is controlled only by you, and not by the local scope.

List * initList()
{
  return malloc(sizeof(List)); // this memory is permanent
}

Any manual allocation needs to come with a clean-up routine:

void freeList(List * p)
{
  free(p);
}

answered Oct 2, 2011 at 0:05

Kerrek SB

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2 Comments

cdn88 Over a year ago

Thank you very much! I was neglecting the scope of List *L! And I am really sorry for making the mistake in my post and wasting peoples time. Take care.

Kerrek SB Over a year ago

No worries, happens to the best :-)

paxdiablo · Accepted Answer · 2011-10-02 07:01:39Z

4

Now that you've posted the remainder of your code, the problem is obvious:

List *callInitialize () {
    List *L;

    List studentList;
    L = &studentList;

    Initialize(L); 

    return L;
}

Within this function you create a local variable studentList which you the pass to Initialize to set it up.

Unfortunately, the scope of the studentList variable ends when you exit the callInitialize function. That's why it may contain rubbish when you attempt to use it later on.

Despite your comment about not needing malloc because the structure is small, you do need it if you want the scope of the data to exist beyond the function it's created in.

Probably the "minimal-change" solution would be:

List *callInitialize (void) {
    List *L = malloc (sizeof (List));
    if (L != NULL)
        Initialize(&L); 
    return L;
}

and then remember that it needs to be freed at some point by the caller, unless the malloc fails and this function therefore returns NULL.

edited Oct 2, 2011 at 7:01

answered Oct 1, 2011 at 23:21

paxdiablo

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1 Comment

cdn88 Over a year ago

I am very sorry I forgot the function callInitialize I added it to the post. I didn't think I would need malloc because my struct items are a set size and not very large.

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