c++,constructor , How does a constructor choose a base class constructor in C++

How does a constructor choose a base class constructor in C++


Tag: c++,constructor

If you are inside the constructor of a derived class, and you make no explicit call to the base class constructor, how does the compiler know which base constructor to use?


If a base class is not mentioned in the constructor initializer list, it will be default initialized. Since the base class will definitely be of class type, that means that the default constructor will be called.

Two of those references also have examples of derived classes which implicitly call the base class default constructor. For instance:

struct Class : public Base
    unsigned char x;
    unsigned char y;

    Class ( int x )
      : Base ( 123 ), // initialize base class
        x ( x ),      // x (member) is initialized with x (parameter)
        y { 0 }       // y initialized to 0
    {}                // empty compound statement

    Class ( double a )
      : y ( a+1 ),
        x ( y ) // x will be initialized before y, its value here is indeterminate
    {} // base class constructor does not appear in the list, it is
       // default-initialized (not the same as if Base() were used, which is value-init)



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