C++直接初始化字段与默认构造函数中的初始化列表

Question

我想知道这段代码是否有区别:

class Foo{
 private:
    int a = 0;
 public:
    Foo(){}
}

还有:

class Foo{
 private:
    int a;
 public:
    Foo(): a(0) {}
}

如果是这样，应该首选哪个？ 我知道使用初始化列表比在构造函数主体中分配更好，但是初始化列表与在字段声明中直接初始化(至少对于原始类型，就像这里的情况一样)呢？

还有，下面是什么情况:

class Foo{
 private:
   int a = 0;
 public:
   Foo(){}
   Foo(int i): a(i) {}
}

当非默认构造函数被调用时:“a”被初始化了两次，先为0然后为“i”，还是直接为“i”？

Answer 1

来自 cppreference - Non-static data members

Member initialization
1) In the member initializer list of the constructor.
2) Through a default member initializer, which is simply a brace or equals initializer included in the member declaration, which is used if the member is omitted in the member initializer list.

If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.

---

总之，这两个初始化器是等价的，并且做它们应该做的事情。

如果我仍然使用默认构造函数，或者如果所有或大多数构造函数将成员初始化为相同的值，我更喜欢默认成员初始化器。

class Foo {
private:
    int a = 0;
};

但是，如果所有构造函数都将成员初始化为不同的值，则使用默认成员初始值设定项意义不大，然后在各自的构造函数中进行显式初始化会更清晰

class Foo {
private:
    int a;
public:
    Foo() : a(3) {}
    Foo(int i) : a(i) {}
};