How to initialize a global array that holds pointers to structs?

Question

I'm brand new to C and am trying to create a global array that will hold pointers to structs:

person* persons[n];

However, I need to make the above a global variable. To do that, I need to declare it outside of all functions. But how can I do that without knowing what n will be? Really lost here, any help would be appreciated!

The below is what I'm ultimately trying to accomplish:

person* persons[];

function () {
    initialize global array here
}

ouah · Accepted Answer · 2015-10-20 21:38:16Z

3

You use a pointer of a pointers and call malloc in an init function.

person **persons;
int n = 42;

void init(void)
{
    persons = malloc(n * sizeof *persons);
    if (!persons) {
        /* handle malloc failure here */
    }
}

answered Oct 20, 2015 at 21:38

ouah

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6 Comments

Vimzy Over a year ago

Can I still use the indices to refer to structs from this new pointer of pointers (as I would in a regular array that holds pointers)?

ouah Over a year ago

@Vimzy yes, for example persons[0] is the first person * of the array object.

Vimzy Over a year ago

@Nevermind, that was my bad. My only problem now is can I still do code like this: "persons[h] = (person *) malloc(sizeof(person))" I'm a little new to this "pointer to pointer" thing with structs.

Vimzy Over a year ago

Because ultimately my code needs this: "persons[h]->name = .." but I'm not sure if this will work?

ouah Over a year ago

@Vimzy yes it will but be sure to initialize all the persons[i] first using malloc as well.

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