如何在C++中的另一个类中声明一个类的构造函数

Question

For example:

class A
{
    public:
    A();
    int a;
};

class B
{
    A::A()
    {
        a = 1;
    }
};

Answer 1

The best way to answer this is in a single word: Don't. In your example, there is no relationship between A and B , so there is no reason for you to even try to define the constructor in another class. The only reason that you would want to mention the constructor of A in B is if B is derived from A . You want to keep parts of A with A , not separate them into other classes and make it a nightmare for readability and maintenance. The short version is that you should keep all parts of A with A , so don't try to put them somewhere else.

Answer 2

No, you can't.

Quoted from the standard [class.mfct] paragraph 1 (emphasized mine):

A member function may be defined in its class definition, in which case it is an inline member function, or it may be defined outside of its class definition if it has already been declared but not defined in its class definition. A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition.

In your example, the definition appear in a class scope, which contradicts to this rule.