struct B {
int b1, b2;
B(int, int);
};
struct D : B {
int d1, d2;
// which is technically better ?
D (int i, int j, int k, int l) : B(i,j), d1(k), d2(l) {} // 1st Base
// or
D (int i, int j, int k, int l) : d1(k), d2(l), B(i,j) {} // last Base
};
Above is just a pseudo code. In actual, I wanted to know that does the order of calling base constructor matter?
Are there any bad behaviors (especially corner cases) caused by any of the cases? My question is on more technical aspect and not on coding styles.
The order you refer in your question is not the "order of calling base constructor". In fact, you can't call a constructor. Constructors are not callable by the user. Only compiler can call constructors.
What you can do is to specify initializers. In this case (constructor initializer list) you are specifying initializers for subobjects of some larger object. The order in which you specify these initializers does not matter: the compiler will call the constructors in a very specific order defined by the language specification, regardless of the order in which you specify the initializers. The base class constructors are always called first (in the order in which the base classes are listed in class definition), then the constructors of member subobjects are called (again, in the order in which these members are listed in the class definition).
(There are some peculiarities in this rule when it comes to virtual base classes, but I decided not to include them here.)
As for the bad behaviors... Of course there is a potential for "bad behaviors" here. If you assume that the order of initialization depends on the order you used in the constructor initializer list, you will most likely eventually run into an unpleasant surprise, when you discover that the compiler completely ignores that order and uses its own order (the order of declaration) instead. For example, the author of this code
struct S {
int b, a;
S() : a(5), b(a) {}
};
might expect a
to be initialized first, and b
to receive the initial value of 5
from a
, but in reality this won't happen since b
is initialized before a
.