是否可以在运行期间在结构中设置数组的大小？

Question

I have a structure as below:

struct Query {  
    int     pages[];
    int     currentpage;
};

I was wondering if it was possible to set the size of this array after creating the structure.

Query new = malloc(sizeof(struct Query));

After this, I will perform some calculations which will then tell me the size that pages[] needs to be. If pages[] needed to be of size 4, how can I set it as such?

Answer 1

In C99 you can use Flexible array members :

struct Query {  
    int currentpage;
    int pages[]; /* Must be the last member */
};

struct Query *new = malloc(sizeof(struct Query) + sizeof(int) * 4);

Answer 2

Change the type of pages member to pointer.

struct Query {  
    int *pages;
    int currentpage;
};

struct Query *test = malloc(sizeof(struct Query));

if (test != NULL)
{
   //your calculations

   test->pages = malloc(result_of_your_calcs);
   if (test->pages != NULL)
   {
      // YOUR STUFF
   }
   else
   {
      // ERROR
   }
}
else
{
   // ERROR
}

When you'll free your struct, you have to do that on the contrary.

free(test->pages);
free(test);

Answer 3

You can use a Flexible array member (details in @AlterMann's answer ) ( C99+ ), or a Zero length array ( GNU C ).

Quoting from https://gcc.gnu.org/onlinedocs/gcc/Zero-Length.html ,

Zero-length arrays are allowed in GNU C. They are very useful as the last element of a structure that is really a header for a variable-length object:

 struct line { int length; char contents[0]; }; struct line *thisline = (struct line *) malloc (sizeof (struct line) + this_length); thisline->length = this_length; 

For standard C90, the linked site mentions

In ISO C90, you would have to give contents a length of 1, which means either you waste space or complicate the argument to malloc .

This means that for the code to work in standard C90/C89, char contents[0]; should be char contents[1]; .

Answer 4

The best solution would be to use pointers to int instead of array.

You will need to change :

int pages[];

to :

int *pages;

and then dynamically allocate it like this :

Query *new = malloc(sizeof(struct Query));
if (new == NULL)
    printf ("Error\n");
else
{
    new->pages = malloc(4*sizeof(int));
    if (new->pages == NULL)
       printf ("Error\n");
}

---

Otherwise if you want to keep your format, you will use C99 mode. Declare pages as the last member of your struct, like this :

struct Query {  
    int currentpage;
    int pages[];
};

and then do :

Query *new = malloc(sizeof(struct Query) + 4*sizeof(int));

Answer 5

Declare it as pointer and use malloc afterwards

struct Query {  
    int * pages;
    int   currentpage;
};

. . .

struct Query obj;
obj.pages = malloc(n * sizeof(int));   // n is the length you want