C++ 默认成员初始化和构造函数

Question

我想知道在哪里可以找到有关以下行为的文档：

class Foo {
public:
  Foo(int argX) : Foo(argX, defaultYValue) {}
  Foo(int argX, int argY) : x(argX), y(argY) {};
private:
  const int x;
  const int y;
  const int defaultYValue = -1;
}

y 值可能未定义吗？或者标准中是否有一些文档说明它有效（我确实注意到默认成员初始化被丢弃，如果它在构造函数中被其他方式覆盖）

PS：这是在忘记默认 YValue 的静态时发现的。

Answer 1

是的，代码有未定义的行为。当使用 delegating constructor 时，委托构造函数将初始化 class 成员。当您将 defaultYValue 传递给委托构造函数时，它尚未初始化，因此您将未初始化的值传递给委托，并且所述委托使用该值来设置 y.

这是由[class.base.init]/7

叫出来的

The expression-list or braced-init-list in a mem-initializer is used to initialize the designated subobject (or, in the case of a delegating constructor, the complete class object) according to the initialization rules of [dcl.init] for direct-initialization.

Answer 2

I would like to know where I can find some documentation about the following behavior:

这称为委托，构造函数Foo::Foo(int)称为delegating constructor。

来自class.base.init-6

A mem-initializer-list can delegate to another constructor of the constructor's class using any class-or-decltype that denotes the constructor's class itself. If a mem-initializer-id designates the constructor's class, it shall be the only mem-initializer; the constructor is a delegating constructor, and the constructor selected by the mem-initializer is the target constructor. The target constructor is selected by overload resolution. Once the target constructor returns, the body of the delegating constructor is executed. If a constructor delegates to itself directly or indirectly, the program is ill-formed, no diagnostic required.

Might it be possible that y value is undefined ?

是的，程序有未定义的行为。

来自class.base.init-7：

The expression-list or braced-init-list in a mem-initializer is used to initialize the designated subobject (or, in the case of a delegating constructor, the complete class object) according to the initialization rules of [dcl.init] for direct-initialization.

这意味着目标构造函数 Foo::Foo(int, int)用于初始化数据成员。但是由于您传递的数据成员 defaultYValue 尚未初始化，因此我们有未定义的行为。

Answer 3

正如您在使用 defaultYValue 时在其他答案中指出的那样，它尚未初始化。您可以通过将其设置为静态来解决此问题：

class Foo {
public:
  Foo(int argX) : Foo(argX, Foo::defaultYValue) {}
  Foo(int argX, int argY) : x(argX), y(argY) {};
private:
  const int x;
  const int y;
  static const int defaultYValue = -1;
}

但是当一个构造函数就足够时，永远不要写 2 个：

class Foo {
public:
  Foo(int argX, int argY = defaultYValue) : x(argX), y(argY) {};
private:
  const int x;
  const int y;
  static const int defaultYValue = -1;
}

但真的为什么要定义一个只用过一次的常量呢？从使用中可以清楚地看出它是 y 的默认值。将其命名为 defaultYValue:

没有获得额外的信息

class Foo {
public:
  Foo(int argX, int argY = -1) : x(argX), y(argY) {};
private:
  const int x;
  const int y;
}

C++ 默认成员初始化和构造函数

C++ default member initialization and constructors

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