std::optional - construct empty with {} or std::nullopt?
In this case, {}
invokes value-initialization. If optional
's default constructor is not user-provided (where "not user-provided" means roughly "is implicitly declared or explicitly defaulted within the class definition"), that incurs zero-initialization of the entire object.
Whether it does so depends on the implementation details of that particular std::optional
implementation. It looks like libstdc++'s optional
's default constructor is not user-provided, but libc++'s is.
For gcc, the unnecessary zeroing with default initialization
std::optional<Data> default_init() {
std::optional<Data> o;
return o;
}
is bug 86173 and needs to be fixed in the compiler itself. Using the same libstdc++, clang does not perform any memset here.
In your code, you are actually value-initializing the object (through list-initialization). It appears that library implementations of std::optional have 2 main options: either they default the default constructor (write =default;
, one base class takes care of initializing the flag saying that there is no value), like libstdc++, or they define the default constructor, like libc++.
Now in most cases, defaulting the constructor is the right thing to do, it is trivial or constexpr or noexcept when possible, avoids initializing unnecessary things in default initialization, etc. This happens to be an odd case, where the user-defined constructor has an advantage, thanks to a quirk in the language in [decl.init], and none of the usual advantages of defaulting apply (we can specify explicitly constexpr and noexcept). Value-initialization of an object of class type starts by zero-initializing the whole object, before running the constructor if it is non-trivial, unless the default constructor is user-provided (or some other technical cases). This seems like an unfortunate specification, but fixing it (to look at subobjects to decide what to zero-initialize?) at this point in time may be risky.
Starting from gcc-11, libstdc++ switched to the used-defined constructor version, which generates the same code as std::nullopt. In the mean time, pragmatically, using the constructor from std::nullopt where it does not complicate code seems to be a good idea.
The standard doesn't say anything about the implementation of those two constructors. According to [optional.ctor]:
constexpr optional() noexcept;
constexpr optional(nullopt_t) noexcept;
- Ensures:
*this
does not contain a value. - Remarks: No contained value is initialized. For every object type
T
these constructors shall beconstexpr
constructors (9.1.5).
It just specifies the signature of those two constructors and their "Ensures" (aka effects): after any of those constructions the optional
doesn't contain any value. No other guarantees are given.
Whether the first constructor is user-defined is implementation-defined (i.e depends on the compiler).
If the first constructor is user-defined, it can of course be implemented as setting the contains
flag. But a non-user-defined constructor is also compliant with the standard (as implemented by gcc), because this also zero-initialize the flag to false
. Although it does result in costy zero-initialization, it doesn't violate the "Ensures" specified by the standard.
As it comes to real-life usage, well, it is nice that you have dug into the implementations so as to write optimal code.
Just as a side-note, probably the standard should specify the complexity of those two constructors (i.e O(1)
or O(sizeof(T))
)