是否有必要在构造函数初始化列表中调用默认构造函数？

Question

例如，如果我有以下内容：

class Foo;  // has default ctor

class Bar {
 public:
  Bar(Foo* f);
}

class MyClass {
 public:
  MyClass();

 private:
  Foo foo_;
  Bar bar_;
}

MyClass::MyClass() : foo_(), bar_(&foo) {}

这或多或少地在没有警告的情况下进行编译，请参见https://godbolt.org/g/yx464A[稍作修改，使 Foo 充实]。

它也可以在初始化列表中没有 foo_() 的情况下正常编译。但是有没有需要调用默认构造函数的情况？

Answer 1

嗯，foo_ 将被默认初始化：

[C++11: 12.6.2/8]: In a non-delegating constructor, if a given non-static data member or base class is not designated by a mem-initializer-id (including the case where there is no mem-initializer-list because the constructor has no ctor-initializer) and the entity is not a virtual base class of an abstract class (10.4), then

if the entity is a non-static data member that has a brace-or-equal-initializer, the entity is initialized as specified in 8.5;

otherwise, if the entity is a variant member (9.5), no initialization is performed;

otherwise, the entity is default-initialized (8.5).

这是否是您想要的取决于您。对于 POD，默认初始化（不会对 POD 成员做任何事情）可能还不够。对于具有要初始化的非 POD 成员的实际默认构造函数的 class，初始化将按照您的预期进行级联。

是否有必要在构造函数初始化列表中调用默认构造函数？

Is it ever necessary to invoke default ctor in constructor initialization list?

c++

initializer-list

c++11