The std::optional<T>::value_or(U &&default_value) function returns the
contained value if available, or default_value if the std::optional has
no value. If the desired default value is a default-constructed T, the
obvious option is to call std::optional<T>::value_or(T{}). This approach
has two drawbacks:
- The \a default_value T{} is constructed even if the std::optional
instance has a value, which impacts efficiency.
- The T{} default constructor needs to be spelled out explicitly in the
value_or() call, leading to long lines if the type is complex.
Introduce a defopt variable that solves these issues by providing a
value that can be passed to std::optional<T>::value_or() and get
implicitly converted to a default-constructed T.
Signed-off-by: Laurent Pinchart <laurent.pinchart@ideasonboard.com>
Reviewed-by: Umang Jain <umang.jain@ideasonboard.com>
Reviewed-by: Jacopo Mondi <jacopo@jmondi.org>
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