CODEWITHSUNDEEP
cstruct
Mazzy
Mazzy

Reputation: 14179

Initialization array in a struct

I don't understand why I got an error when I try to initialize a struct in such a way

typedef struct _floor
{
   int room;
   int height;
   int room_dim[room][2];
}Floor;

Why can't I use room to initialize room_dim array?

Upvotes: 1

Views: 180

Answers (3)

Madan Ram
Madan Ram

Reputation: 876

you can instead use malloc to get the dynamic memory rather then using room_dim[room][2] .Since the above is not the allowed standard in c compiler.

for example

typedef struct floor
{
   int *room_dim[];
   int height;
   int room;
}floor;
scanf("%d",&room);
floor.room_dim=(floor *) malloc(sizeof(floor)*room);

Upvotes: 1

nneonneo
nneonneo

Reputation: 179382

A struct must have a size that is known at compile-time. room is a struct variable, and could have any value; therefore, it is not a compile-time constant and cannot be used to size a struct member.

Instead, you can make the final element a flexible array member and allocate it at runtime:

struct floor {
    int rooms;
    int height;
    int room_dim[][2];
};

struct floor *make_empty_floor(int rooms) {
    struct floor *ret = malloc(sizeof(struct floor) + sizeof(ret->room_dim)*rooms);
    ret->rooms = rooms;
    return ret;
}

Now you can use ret->room_dim as usual, and the extra malloc'd space will be used for room_dim.

Upvotes: 5

user2618142
user2618142

Reputation: 1065

You are trying to dynamically initialize an array. This means you will only know the size of room at run time.

[Edited after comments]

To understand it simply, without use of struct, even this won't compile below C99, however in structs this is still not allowed

int n;  
int size[n];

Upvotes: 1

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