In the following code, when the ctor of X
is called will the ctor of A
or B
be called first? Does the order in which they are placed i
The order is the order they appear in the class definition - this is from section 12.6.2 of the C++ Standard:
5 Initialization shall proceed in the following order:
— First, and only for the constructor of the most derived class as described below, virtual base classes shall be initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base class names in the derived class base-specifier-list.
— Then, direct base classes shall be initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).
— Then, nonstatic data members shall be initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).
— Finally, the body of the constructor is executed. [Note: the declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. ]
Initialization is always in the order that the class members appear in your class definition, so in your example a
, then b
.
There is a sequence point between the initialization of each member and you can pass a reference to a yet-to-be initialized member into the constructor of a class member but you would only be able to use it in limited ways (such as taking its address to form a pointer), other uses may well cause undefined behaviour.
Destruction of class members always happens in the reverse order of construction.
Order of initialization of bases and members is defined in 12.6.2 [class.base.init]/5.