c++ - 类内成员初始化器与初始化列表的冲突解决

Question

让我在问题之前添加以下代码:

struct A
{
  explicit A(int i):a_{i} {}

  int a_
};

struct B
{
  B():mya_{5} {} // Initialize mya_ (again?)

  A mya_{7}; // Initialize mya_
};

在struct B我们在 mya_ 的类内初始值设定项之间存在冲突和 mya_在 B 中初始化的构造函数的初始化列表。根据 C++ 标准如何解决这个问题，mya_.a_ 的最终值应该是多少？什么时候B搭建完成了吗？

Answer 1

初始化列表获胜。如果您有另一个不初始化成员的构造函数，就地初始化将获胜。

例如，

struct B
{
  B():mya_{5} {}
  B(int) {}
  A mya_{7};
};

int main
{
  B b0;    // b.mya_.a_ is 5
  B b(42); // b.mya_.a_ is 7
}

来自12.6.2 初始化基类和成员[class.base.init]

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 ]