在Struct内部创建字符串数组

Question

I'm trying to create an array of strings inside a struct which represents a player's inventory.

I create a struct for the player:

typedef struct Player {
    char *inventory[];
} Player;

And then I use a function which allocates heap memory and creates (not sure if "create" is the right word here) the struct with some "items" inside the player's "inventory".

Player *spawnPlayer(void)
{
    Player *stats = malloc(sizeof(Player));

    stats->inventory[] = {"potion", "potion", "ether"};

    return stats;
}

Now I can create a normal array like this outside of a struct, but if I attempt to use the above, I get the below error while trying to compile:

arrays.c: In function 'spawnPlayer':
arrays.c:13:19: error: expected expression before ']' token
  stats->inventory[] = {"potion", "potion", "ether"};

Would someone be able to point me in the right path as to why this doesn't work?

Answer 1

Your definition of Player is an example of a C99 extension call flexible array: an array of unknown size that is the last member of a structure and that will be accessible only upto the size actually allocated for each instance. You probably did not mean to use that. And you cannot initialize the values with the syntax used in your spawnPlayer function.

You can define an array of fixed size this way:

typedef struct Player {
    char *inventory[3];
} Player;

And you can initialize an allocated instance of Player this way:

Player *spawnPlayer(void) {
    Player *stats = malloc(sizeof(*stats));

    stats->inventory[0] = "potion";
    stats->inventory[1] = "potion";
    stats->inventory[2] = "ether";

    return stats;
}

If you meant for the array to have a size known at runtime and use a flexible array, you probably want to add a member for the actual size allocated:

typedef struct Player {
    int size;
    char *inventory[];
} Player;

And you will allocate it and initialize it this way:

Player *spawnPlayer(void) {
    Player *stats = malloc(sizeof(*stats) + 3 * sizeof(*stats->inventory));

    stats->size = 3;
    stats->inventory[0] = "potion";
    stats->inventory[1] = "potion";
    stats->inventory[2] = "ether";

    return stats;
}

Flexible arrays are a C99 extension, you can simulate them in C90 by defining inventory with a size of 0 is the compiler supports it, or a size of 1 , but it is not strictly portable.

There is a third possibility, using a pointer to an array of char* :

typedef struct Player {
    int size;
    char **inventory;
} Player;

Player *spawnPlayer(void) {
    Player *stats = malloc(sizeof(*stats));

    stats->size = 3;
    stats->inventory = malloc(sizeof(*stats->inventory) * 3);
    stats->inventory[0] = "potion";
    stats->inventory[1] = "potion";
    stats->inventory[2] = "ether";

    return stats;
}

Answer 2

typedef struct Player {
    char *inventory[10];
} Player;

static const char *arr[] = { "potion", "potion", "ether"};
memcpy(stats->inventory, arr, sizeof(arr));

Or

typedef struct Player {
    char **inventory;
} Player;

Player *stats = malloc(sizeof(Player));
stats->inventory = (char**)malloc(sizeof(char*) * 10);

static const char *arr[] = { "potion", "potion", "ether"};
memcpy(stats->inventory, arr, sizeof(arr));

Or

stats->inventory[0] = "potion";
stats->inventory[1] = "potion";
stats->inventory[2] = "ether";

Perhaps you should add a NULL pointer at the end to indicate the length.

stats->inventory[3] = NULL;

static const char *arr[] = { "potion", "potion", "ether", NULL };
memcpy(stats->inventory, arr, sizeof(arr));

Or simply have a int inventory_size in your struct.

Answer 3

You need to explicitly declare an array called inventory specifying its size.

Then initialize a pointer to point to it.

This is the third time I've seen this mistake in two days.