Why could const member be initialized twice?

Question

Below is a code snippet which could be compiled and run without error in vs2015

#include<iostream> using namespace std;  class A {     public:         A(int b) :k(b) {}//second time     const int k = 666;//first time };  int main() {     A a(555);     cout << a.k << endl;     return 0; } 

The output is 555. But as far as I know,const object should be initialized only once,after which the value is unmodifiable.

Answer 1

It's not initialized twice; the default member initializer is just ignored. So for A a(555);, a.k is initialized as 555.

If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.

From the standard, [class.base.init]/10:

If a given non-static data member has both a default member initializer and a mem-initializer, the initialization specified by the mem-initializer is performed, and the non-static data member's default member initializer is ignored. [ Example: Given

struct A {   int i = /* some integer expression with side effects */ ;   A(int arg) : i(arg) { }   // ... }; 

the A(int) constructor will simply initialize i to the value of arg, and the side effects in i's default member initializer will not take place. — end example ]

On the other hand, given

class A { public:     A() {}            // k will be initialized via default member initializer, i.e. 666     A(int b) :k(b) {} // k will be initialized via member initializer list, i.e. b      const int k = 666; }; 

then for A a;, a.k will be initialized as 666.

Answer 2

It is initialized only once.

const int k = 666; 

would be used if not provided in constructor.