Predict the output of following C++ program:
B's Constructor Called
The above program calls only B’s constructor, it doesn’t call A’s constructor. The reason for this is simple, static members are only declared in class declaration, not defined. They must be explicitly defined outside the class using scope resolution operator.
If we try to access static member ‘a’ without explicit definition of it, we will get compilation error. For example, following program fails in compilation.
Compiler Error: undefined reference to `B::a'
If we add definition of a, the program will works fine and will call A’s constructor. See the following program.
A's constructor called B's constructor called B's constructor called B's constructor called
Note that the above program calls B’s constructor 3 times for 3 objects (b1, b2 and b3), but calls A’s constructor only once. The reason is, static members are shared among all objects. That is why they are also known as class members or class fields. Also, static members can be accessed without any object, see the below program where static member ‘a’ is accessed without any object.
A's constructor called
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
- Initialization of data members
- Can we access private data members of a class without using a member or a friend function?
- Understanding "static" in "public static void main" in Java
- Difference between static and non-static variables in Java
- Difference between static and non-static method in Java
- Different ways to Initialize all members of an array to the same value in C
- Flexible Array Members in a structure in C
- Are array members deeply copied?
- C++ Program to swap two members using Friend Function
- Static Keyword in C++
- Static functions in C
- Static Variables in C
- When are static objects destroyed?
- C++ | Static Keyword | Question 2
- C++ | Static Keyword | Question 3