c++ - 使用构造函数与大括号初始化列表来初始化类和结构的规则是什么？

Question

我已经在网上搜索了这个问题的答案，但我还没有找到满意的答案。我想知道结构和类类型对象初始化的所有规则是什么，特别是在构造函数与大括号初始化列表时。结构与类的规则也不同吗？

让我们假设我们有一个名为 的类或结构Rectangle。

#include <iostream>
using namespace std;

class Rectangle {
  public:
    Rectangle() : x(5.0), y(6.0), width(7.0), height(8.0) {}

    void printMe()
    {
        cout << "The rectangle is located at (" << x << ',' << y << ") and is " << width << " x " << height << endl;
    }
    double x;
    double y;
    double width;
    double height;
};


int main()
{
    Rectangle r = {0.0, 0.0, 3.0, 4.0};
    r.printMe();

    Rectangle s;  // uninitialized!
    s.printMe();
}

我尝试Rectangle r使用普通的旧大括号初始化器列表来初始化您通常在 C 中执行的方式。但是，g++给出以下错误：

constructor_vs_initializer_list.cpp: In function ‘int main()’:
constructor_vs_initializer_list.cpp:21:38: error: could not convert ‘{0.0, 0.0, 3.0e+0, 4.0e+0}’ from ‘<brace-enclosed initializer list>’ to ‘Rectangle’
     Rectangle r = {0.0, 0.0, 3.0, 4.0};
                                      ^

嗯....乍一看，这不是很有用的错误消息。但是我认为它与构造函数有关，因为如果我删除它，代码就会编译并运行！我认为这将是一个悖论，大括号初始化列表和构造函数都在竞争初始化它看起来的数据成员。

但是，当我制作数据成员private时，删除构造函数后，再次显示相同的错误消息！

我想知道数据成员初始化的优先级规则是什么。大括号初始值设定项列表与您自己定义的构造函数相比如何？它与 C++11 的特性相比如何：= default构造函数和类内成员初始化器？我假设这些初始化对象数据成员的不同方式会以某种方式相互冲突。

Rectangle() = default;
...
double x = 1.0;

我并不是说将它们混合在一起一定是好的代码，只是它是代码，并且在我看来应该很好理解的代码。谢谢你。

标签： c++c++11constructorinitializer-listlist-initialization

Answer 1

这是一个演示差异的示例。C++ 中的初始化非常复杂。请参阅：https ://blog.tartanllama.xyz/initialization-is-bonkers/ 。

通常最好使用默认成员初始化器或初始化列表。你在你的构造函数中做正确的事情。只需使用直接列表初始化或直接初始化调用构造函数，以避免混淆。通常，您只会使用复制列表初始化来初始化聚合，而无需用户提供的构造函数。

#include <iostream>

struct A {
    int i;
};

struct B {
    B() = default;
    int i;
};

struct C {
    C();
    int i;
};

C::C() = default;

struct D {
    D(){};
    int i;
};

struct E : public D {
};

struct F {
    F(int i = 5) {}
    int i;
};

struct G {
    G() = delete;
    int i;
};

int main() {
    // g++ (v 8.2.1) provides good warnings about uninitialized values.
    // clang++ (v 7.0.1) does not.
    // Technically, they are initialized to 'indeterminate values', but it is
    // easier to refer to the member variables as uninitialized.

    {
        // All of the following are 'default initialized', meaning they are not even
        // zero-initialized. Members are UNINITIALIZED (Technically, they are
        // initialized to 'indeterminate' values.
        // Either nothing is done, or the default constructor is called (in
        // which nothing is done).
        A a;
        B b;
        C c;
        D d;
        E e;
        F f;

        std::cout << "a: " << a.i << std::endl;
        std::cout << "b: " << b.i << std::endl;
        std::cout << "c: " << c.i << std::endl;
        std::cout << "d: " << d.i << std::endl;
        std::cout << "e: " << e.i << std::endl;
        std::cout << "f: " << f.i << std::endl;
        std::cout << std::endl;
    } {
        // This is more complex, as these are all 'list initialized'.
        // Thank you, infinite wisdom of the C++ committee.

        A a{};
        // Direct list initialization -> aggregate initialization
        //  - A has no user-provided constructor and
        // thus is an aggregate, and agg. init. takes place.
        // This 'value initializes' all *MEMBERS* (unless a default member
        // initializer exists, which it does not here).
        // Value initialization of non-class types results in
        // zero-initialization. (member `i` is zero-initialized)

        A a2 = {};
        // same thing, but via copy list initialization

        A a3{{}};
        // recursive, initializes `i` with {}, which zero initializes `i`.

        A a4{0};
        // recursive, initializes `i` 0;
        // Could also do `A a4 = {0}`

        A a5{a};
        // direct intialization of `a5` with `a`.
        // Implicit copy constructor chosen by overload resolution.

        A a6{A{}};
        // post C++17, direct initializes a6 with a prvalue of type A, that is
        // aggregate initialized as above. NOT copy/move initialized, but
        // instead initialized via the "initializer expression itself".
        // I assume this means the value of a6 is directly set via as if it were
        // being aggregate initialized.

        B b{};
        // Same as A. `B() = default;` does NOT specify a user-provided
        // constructor

        C c{};
        // Because the first declaration of `C()` is not `C() = default;`,
        // this DOES have a user-provided constructor, and 'value initializaton'
        // is performed.
        // NOTE: this value intializes `C`, not the *MEMBERS* of `C`.
        // Because `C` is a normal class type, value initialization just calls
        // the default constructor, which does nothing, and leaves all members
        // uninitialized.

        D d{};
        // D is a class type that is list/direct initialization -> value
        // inititalizaton -> default initialization -> call constructor ->
        // members are left unitialized.

        E e{};
        // List initialization -> value initialization -> default initialization
        // -> calls implicitly defined default constructor -> Calls default
        // constructor of bases -> leaves E::D.i uninitialized

        F f{};
        // List/direct initialization -> value initialization -> calls default
        // constructor with default arguments -> leaves F.i uninitialized

        // G g{}; 
        // Fails to compile.
        // list initialization -> value initialization -> default initialization
        // -> deleted default constructor selected by overload resolution ->
        // fails to compile

        std::cout << "a: " << a.i << std::endl;
        std::cout << "a2: " << a2.i << std::endl;
        std::cout << "a3: " << a3.i << std::endl;
        std::cout << "a4: " << a4.i << std::endl;
        std::cout << "a5: " << a5.i << std::endl;
        std::cout << "a6: " << a6.i << std::endl;
        std::cout << "b: " << b.i << std::endl;
        std::cout << "c: " << c.i << std::endl;
        std::cout << "d: " << d.i << std::endl;
        std::cout << "e: " << e.i << std::endl;
        std::cout << "f: " << f.i << std::endl;
    }
}