类成员初始化程序和初始化列表之间的冲突解决 [英] Conflict resolution between in-class member initializer and initialization list

问题描述

让我先使用下面的代码：

Let me precede the question with the following code:

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 在 B mya _ 和 mya _ c>的构造函数的初始化器列表。如何解决每个C ++标准， mya_.a _ 的最终值应该是 B 的结构是否完成？

In struct B we have a conflict between the in-class initializer for mya_ and mya_ being initialized in B's constructor's initializer list. How is this resolved per the C++ standard and what should the final value of mya_.a_ be when B's construction is complete?

推荐答案

初始化列表获胜。

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
}

从 12.6.2正在初始化基础和成员[class.base.init]

From 12.6.2 Initializing bases and members [class.base.init]

如果给定的非静态数据成员都有
brace-or-equal-initializer和mem初始化器，执行由mem初始化器指定的初始化
，并且忽略非静态数据
成员的括号或等于初始化器。 [示例：给定

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) { }
// ...
};

A（int）构造函数将简单地将i初始化为arg，
，并且在i的括号或者等号初始化器中的副作用不会取
。 - end example]

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 ]

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