int a;
scanf("%d",a);
typedef struct mylist {
int info[a];
struct mylist *link;
} Node;
this is my very simple struct and variable 'a' that don't work.
The thing which I want my program to do is to read input from the user and insert the digit into the struct (the user can determine the size of the array), however, I don't know how to do it since I get an error:
"variably modified ‘info’ at file scope".
I want to know whether it is possible to do this in a simple way.
When you use scanf into an integer you must pass a pointer. You may want to check the documentation.
scanf("%d", &a);
As far as using a variable to allocate data, you can do it, but you will want to use malloc most likely after defining your structure as a type.
scanf()?
"%d" specifier is used, things can go very wrong if one assumes that it worked!
As pointed out by @David Hoelzer
scanf("%d",a);
should be
scanf("%d",&a);
If you are under C99 you can use flexible array members:
typedef struct mylist {
struct mylist *link;
int info[]; /* should be the last member */
} Node;
and then
Node *node = malloc(sizeof *node + (a * sizeof(int)));
You must have pointer to array of correct size or use flexible array. Here is how to use a pointer initializing it to point to storage of proper size.
struct mylist {
int *info; // or int info[] at the end of struct def if C99 is used
int info_size;
struct mylist *link;
};
int main(void)
{
int *array;
struct mylist m;
if (scanf("%d", &m.info_size) != 1)
{
perror("scanf failed");
exit(EXIT_FAILURE);
}
// ok
array = malloc(sizeof(int) * m.info_size);
if (array == NULL)
{
perror("malloc failed");
exit(EXIT_FAILURE);
}
// ok, commit changes
m.info = array;
scanf("%d", a);should bescanf("%d", &a);int *info;and usemalloc:Node n;,n.info = malloc((sizeof int) * a);.struct.