Question

由于四舍五入,如果你在一类班头上有一个私人成员定义,这种定义就是:

box newBox;

盒子里有一个构造者,不接受任何论点,这是否意味着你必须把它当作点子或参考物?

你们如何处理这个问题? 似乎难以以这种方式宣布某些课堂,而不是接受任何论点的人。或者我是否误解?

我的理解是,如果它不接受任何论点,那么这不仅是一个定义,也是初步的。

class whatever
{
private:
    box newBox; //is this not allowed because of the most vexing parse? because it s a
                //header file and because of the MVP this is initialisation as well? or
                //am I getting confused about something?
};

class box
{
public:
    box();
}

Answer 1

以下著作只是罚款:

struct Foo {
}

struct Bar {
    Foo foo;
}

无地雷影响调查,没有模糊之处。一切都属于罚款。在一类申报中<代码> Foo foo; is a member-declaration. 不可能在该地进行初始化,在构造中(此处为<条码>Bar)进行初步化(明示或默示)。

Answer 2

If the only constructor accepts no arguments, this is the canonical constructor that s called every time you construct an instance of the class which has newBox as an element. It s still always a new box object, though it ll always look the same.

You can test it this way:

class foo {
 public:
  foo() {
    std::cout << "Constructed a foo.
";
  }
};

class bar{
  foo afoo;
 public:
  bar() {}
};

int main() {
  std::cout << "Request new bar: ";
  bar bar1;
  std::cout << "Request another new bar: ";
  bar bar2;
  return 0;
}

产出:

Request new bar: Constructed a foo.
Request another new bar: Constructed a foo.

Answer 3

页: 1 除需要将<代码><> 方框>/代码”的定义移至<代码>whatever的类别外,还可在该栏中采用<代码>。当你创建<代码>whatever时,newBox将与<代码>方框级的违约构造者一起启动。

如果你想明确初始化或以某种方式定制,就必须在<代码>上添加一个构件。班级。例如:

class whatever {
private:
    box newBox;
public:
    whatever() : newBox() {
       // ...
    }
};

Answer 4

这是最令人 v慕的教区的一个例子:

box newBox();

这并不是:

box newBox;

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