malloc与多维数组

Question

I'm working on a dictionary whose structure is:

typedef union _dict {
    union _dict * children[M];
    list * words[M];
} dict;

Initialisation:

dict *d = (dict*) malloc(sizeof(dict));

I'm trying to do the following:

dict *temp;
temp = d;

temp=temp->children[0];
temp=temp->children[0];

The first temp->children[0] works but not the second. I'm trying to understand why. I think it's a memory allocation problem.

Edit 1: I've tried the following code :

dict *d = (dict*) malloc(sizeof(dict));

dict *temp;
temp = d;

dict *d2 = (dict*) malloc(sizeof(dict));
temp->children[0] = d2;

temp = temp->children[0];
temp = temp->children[0];
temp = temp->children[0];

That now works, but I don't understand why... I mean, i don't have allowed some memory for next children.

Edit 2: So now, I would like to use this in my algorithm. The code block where I am stuck is the following:

list *l;
if (temp->words[occur] != NULL) {
    /* ... */
}
else {
    l = list_new();
    temp->words[occur] = (list*) malloc(sizeof(list));
    temp->words[occur] = l;
}
list_append(l,w);
list_print(l);

If I put a temp->words[occur] = NULL; before this block, the word is successfully added, but a new list is created each time the algorith is used. I would like to add my word to the previously created list, assuming it exists.

A bzero((void*)d, sizeof(dict)); instruction is used after the dict initialisation.

Answer 1

At first in temp you have a valid pointer to an object (allocated with malloc). Then you assign an uninitialized pointer to temp and attempt to dereference it with the expected consequences.

Answer 2

children is never initialised, so it contains whatever garbage was in memory before. After the first temp = temp->children[0] , temp is a pointer to unknown territory.

Answer 3

The problem lies in that the first children is created as a pointer to a dict like: union _dict * children[M]; but each element in that array is not allocated automatically. Therefore, your second call is using a "child" that hasn't had dict space created for it.

If you want it to work properly something like this should solve the issue:

typedef union _dict {
    union _dict * children[M];
    list * words[M];
} dict;

dict *temp = (dict*) malloc(sizeof(dict));

//allocate space for child, and reassign to it
temp->children[0] = (dict*) malloc(sizeof(dict));
temp = temp->children[0];

//allocate space for child, and reassign to it
temp->children[0] = (dict*) malloc(sizeof(dict));
temp = temp->children[0];

//etc...

Each child has to be allocated memory, or you just get trash and/or seg faults.

Answer 4

temp=temp->children[0];

You never assign any value for temp->children[0] . So after this line of code, temp contains garbage.

Answer 5

i think 2nd time usage might give you segmantation fault.

see

dict *d = (dict*) malloc(sizeof(dict));  // once you have malloc



dict *temp;
temp = d;

temp=temp->children[0];  // that is going to use here

for using 2nd time you havent malloc..? right so beffre to use in that way you should malloc like

 d->children[0] = (dict*) malloc(sizeof(dict));