CODEWITHSUNDEEP
carraysstructflexible-array-member
user9597158
user9597158

Reputation:

Invalid use of flexible array member (Not as the others)

So I have these 2 structures:

typedef struct item { 
    const char *label;
    int value;
} Item;

typedef struct item_coll { 
    size_t length; 
    Item items[]; 
} ItemColl;

And I want to do this:

 int main() {

    Item a = {"A", 10};
    Item b = {"B", 20};
    Item c = {"C", 30};

    Item items[] = {a, b, c};

    size_t length = sizeof(items)/sizeof(items[0]);

    ItemColl *column = malloc (sizeof(column) + length * sizeof(Item));

    column -> length = length;
    column -> items = items;

    printf("%ld\n", column -> length);

    return 0;
}

But I'm getting the error "Invalid use of flexible array member" here:

column -> items = items;

As far as I know, I'm allocating the needed space, which is why I don't understand what the problem is.

I've seen 2 more posts with this title but none of them solves my problem, as I've tried the answers to those questions.

Upvotes: 0

Views: 1255

Answers (1)

Craig Estey
Craig Estey

Reputation: 33601

As others have mentioned, you can't assign one array to another.

Partly, because the compiler can't always know how long the array is, particularly for a flexible array member. (e.g.) Also, either the source or target might be a pointer. To be consistent, it just flags it.

So, change:

column->items = items;

to:

for (int idx = 0;  idx < length;  ++idx)
    column->items[idx] = items[idx];

Or, to use memcpy:

memcpy(column->items, items, sizeof(column->items[0]) * length);

Side note:

If column->items were a pointer (e.g. Item *items) instead, doing:

column->items = items;

would be valid. But, it would not copy the values. It would just set the struct's pointer to the address of the function scoped array items. This would not be the desired result.

Upvotes: 1

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