...one of the most highly
regarded and expertly designed C++ library projects in the
world.
— Herb Sutter and Andrei
Alexandrescu, C++
Coding Standards
Technical details aside, the memory layout of optional<T>
is more-less this:
template <typename T> class optional { bool _initialized; std::aligned_storage_t<sizeof(t), alignof(T)> _storage; };
But for the purpose of this analysis, considering memory layouts, we can think of it as:
template <typename T> class optional { bool _initialized; T _storage; };
Given type optional<int>
, and
assuming that sizeof(int) ==
4
, we will get sizeof(optional<int>)
== 8
.
This is so because of the alignment rules, for our two members we get the
following alignment:
This means you can fit twice as many int
s
as optional<int>
s into
the same space of memory. Therefore, if the size of the objects is critical
for your application (e.g., because you want to utilize your CPU cache in
order to gain performance) and you have determined you are willing to trade
the code clarity, it is recommended that you simply go with type int
and use some 'magic value' to represent
not-an-int.
Even if you cannot spare any value of int
to represent not-an-int (e.g., because every value is
useful, or you do want to signal not-an-int explicitly),
at least for Trivial
types
you should consider storing the value and the bool
flag representing the null-state separately. Consider
the following class:
struct Record { optional<int> _min; optional<int> _max; };
Its memory layout can be depicted as follows:
This is exactly the same as if we had the following members:
struct Record { bool _has_min; int _min; bool _has_max; int _max; };
But when they are stored separately, we at least have an option to reorder them like this:
struct Record { bool _has_min; bool _has_max; int _min; int _max; };
Which gives us the following layout (and smaller total size):
Sometimes it requires detailed consideration what data we make optional. In our case above, if we determine that both minimum and maximum value can be provided or not provided together, but one is never provided without the other, we can make only one optional memebr:
struct Limits { int _min; int _max; }; struct Record { optional<Limits> _limits; };
This would give us the following layout:
Having function parameters of type const
optional<T>&
may incur certain unexpected run-time cost connected to copy construction
of T
. Consider the following
code.
void fun(const optional<Big>& v) { if (v) doSomethingWith(*v); else doSomethingElse(); } int main() { optional<Big> ov; Big v; fun(none); fun(ov); // no copy fun(v); // copy constructor of Big }
No copy elision or move semantics can save us from copying type Big
here. Not that we need any copy, but
this is how optional
works.
In order to avoid copying in this case, one could provide second overload
of fun
:
void fun(const Big& v) { doSomethingWith(v); } int main() { optional<Big> ov; Big v; fun(ov); // no copy fun(v); // no copy: second overload selected }
Alternatively, you could consider using an optional reference instead:
void fun(optional<const Big&> v) // note where the reference is { if (v) doSomethingWith(*v); else doSomethingElse(); } int main() { optional<Big> ov; Big v; fun(none); fun(ov); // doesn't compile fun(v); // no copy }