Re: Zero-size array as struct member
Bo Persson <bop@gmb.dk> wrote:
No, in C++ you would use a std::vector in place of the MyStruct, not
in place of the buffer pointer. That is what people have been trying
to tell you here.
Care to show me how it would be done?
The idea in the original post, in other words, the idea with the so-called
"struct hack", is that you have an object which has an array as member, the
size of this array being determined at runtime. In other words, what you
would normally write like this:
struct MyStruct
{
int someData, someOtherData;
std::vector<int> anArray;
};
That's a struct which has an array as member. Its minor "problem" is
that the array is allocated separately from the struct itself. The struct
just contains a pointer to that array (which is what std::vector really is,
internally). This usually doesn't matter much. However, if you need to
instantiate MyStruct dynamically, and you need to do a lot, then you are
effectively doubling the amount of allocations being performed (each
allocation of MyStruct also causes an additional allocation performed
by std::vector).
The C struct hack is a low-level optimization of this, where the array
is allocated alongside the struct itself. Due to how C (and C++) works,
this is technically possible and feasible. Thus you write the above as:
struct MyStruct
{
int someData, someOtherData;
int anArray[0];
};
and then when allocating an instance of MyStruct, you "overallocate" memory
for it. The additional memory is used as the memory needed by the 'anArray'
member, which can then be accessed normally with operator[].
This way only one allocation is needed, even though the size of 'anArray'
is determined at runtime.
Now, would you kindly show me how std::vector<MyStruct> achieves this
same result?
I think you are confused about what the struct hack is all about. Please
try to understand what is it that the struct hack is doing. I have tried
to explain it as well as I can.