Peteris Krumins said:
Hello,
C99 Standard states that the last member of structure can be
an array whose size is omitted. What's the point of having
such member in a structure and how can it be effectivly
used?
struct foo {
int value;
long another;
char arr[0];
};
Actually, you didn't omit the size, you made it zero, which is
wrong.
The proper C99 syntax is:
struct foo {
int value;
long another;
char arr[];
};
The way you use it is usually with malloc:
struct foo *p = malloc(n + sizeof *p);
if(p != NULL)
{
/* Now p->arr has n elements available. */
/* ... */
free(p);
}
This was added because of the *wrong* existing practise to do:
struct foo {
int value;
long another;
char arr[1]; /* Note one element */
};
struct foo *p = malloc(n + sizeof *p - 1);
if(p != NULL)
{
/* Now p->arr is used as if it has n elements available. */
/* ... */
free(p);
}
This is wrong because it is still undefined behaviour to
access an element outside the declared array size of 1.
It does work on most C89 compilers.
Also can i simply have a variable (array) which has size
omitted?
Only in a couple of cases. The first dimension of a function
parameter declared using array syntax can have its size
omitted. This is OK because the array size is never used --
the parameter's type is always rewritten into a pointer to
the first element.
For example:
int foo(int arr[][5]);
is rewritten as:
int foo(int (*arr)[5]);
Another example is where you're making a declaration of an
array which is defined in another translation unit:
extern int arr[];
int omitted[0];
char *pointers[0]; /* is this valid too? */
Neither of these are valid. Omitting the size does *not*
mean you can make it zero.
