how does this Implicit conversion happen?

how does this Implicit conversion happen?

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

A a;
B b=a;

class B has a constru开发者_运维技巧ctor B(A a) takes object a as a object,but inside the constructor of B, it does nothing. so how does this happens?

---

The constructor of B which takes a parameter of type A gets called when you write B b=a. Its true that constructor of B does nothing with the argument a, but that is a different thing altogether.

Its equivalent to:

 B b(a); //equivalent to "B b = a;"

---

The syntax B b=a invokes the constructor of B which takes A as argument; after that what the constructor does with the argument a, has nothing to do with the constructor's invocation!

It is like the following case, when you write f(100), you're passing 100 to f(); now it f does nothing with the 100, its a different story. It doesn't has anything to do with the function invocation.

void f(int a)
{
  return;
}

f(100);

---

so how does this happens?

B(A a) gets called. Pretty much the same thing that would happen in that constructor as if b was default constructed with an empty default constructor. Default constructors on members of B are executed.

---

This is happening because in C++, while calling constructor, implicit conversion for one parameter is allowed.

For eg. if you have the following class:

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

Then, the following statement is valid:

A ob = 5 ;
equivalent to `A ob(int(5)) ;`

Drawing analogy from this example, in your case, B b = a; can be expanded as B b(A(a));

Further, you can read about explicit keyword in C++

Note: The example above is just for reference so please ignore the equivalent notations for syntactical validation

---

Conversion from A to B means, "construct an instance of B, passing an instance of A as a constructor parameter".

Whether B actually uses that instance of A to do anything in the constructor is irrelevant. It's still a conversion, whether or not the value of the A instance has any effect on the result.

---

B's constructor doesn't need to do anything.

By creating the [non-explicit] constructor that takes an A as a parameter, you're making A convertible to B.

The constructor body is where you can specify behaviours that should be performed upon such a construction, but since you didn't write anything there (and nothing in the ctor-initializer, either), nothing happens.

The actual construction of the B still occurs, as usual, thanks to the language.

You just get the B, as if you'd created it through any other constructor you could come up with, either without arguments or with arguments that you never get around to using.

^{BTW, you should probably accept A& const, not A.}