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 in the body of the class control this? If somebody can provide a snippet of text from the C++ standard that talks about this issue, that would be perfect.
class A {};
class B {};
class X
{
A a;
B b;
};
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.