On Tue, May 14, 2013 at 9:41 AM, Olivier Langlois
<olivier@xxxxxxxxxxxxxxxxxxx> wrote:
> On Tue, 2013-05-14 at 08:52 +0200, Martti Kühne wrote:
>> Sorry if this is OT and a dumb question, but are you sure you want
>>
>> On Mon, May 13, 2013 at 8:20 PM, LANGLOIS Olivier PIS -EXT
>> <olivier.pis.langlois@xxxxxxxxxxxxxxxxxxxx> wrote:
>> [...]
>> > struct B
>> > {
>> > int numelem;
>> > /*
>> > * Old C trick to define a dynamically sizable array just by allocating
>> > * sizeof(B) + (numelem-1)*sizeof(A) memory.
>> > */
>> > A item[1];
>> > };
>> >
>>
>> one item vs.
>>
> That is a old C trick when STL container did not exist. The other
> options would be to replace the 1 item array with pointer but then you
> would have to do 2 malloc. 1 for struct B 1 for the array of A.
>
> Maybe my usage of B array did obfuscate the pattern. An another way to
> use it is:
>
> B *p = (B *)malloc(sizeof(B) + (numelem-1)*sizeof(A));
>
> That way you can define dynamically the say of the 'item' array. This
> pattern dates back way before C99 standard which did, I think, introduce
> dynamic array where you could write
>
> B array[var];
>
>
>
C99 variable-length arrays are put on the stack, they're just a
standard alloca. The standard/valid way of having a flexible-length
struct member is `array[]` without a given size - no workaround is
needed.
