大括号或等于初始化程序和初始化列表之间的 C++ 评估顺序？

Question

我有这个结构的地方，

struct
AAA
{
    AAA() : bbb(2)
    {
        // ccc ???
    }

    int bbb = 1;
    int ccc = bbb;
};

AFAIK ，如果有一个初始化列表 :bbb(2)，表达式 bbb = 1 将被忽略。然后，我不清楚 ccc 最终会变成什么。

initialization-list 或 brace-or-equal initializer 中的哪一个会首先被评估？它们之间的规则是什么？

Answer 1

C++11 draft §12.6.2.9 说；

If a given non-static data member has both a brace-or-equal-initializer and a mem-initializer, the initialization specified by the mem-initializer is performed, and the non-static data member’s brace-or-equal-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 brace-or- equal-initializer will not take place. — end example ]

由于添加了这条规则，初始化是按照声明顺序(§12.6.2.10)完成的，bbb 和 ccc 的值都将是 2。