问题描述
想象一下这个类:
class Entity {
public:
int x, y;
Entity() : x(0), y(0) { }
Entity(int x, int y) : x(x), y(y) { }
}
这里有多种使用我认为我知道的初始化类的方法:
And here are multiple ways of initializing the class with what I think I know:
Entity ent1; //Uses the default constructor, so x=0 and y=0
Entity ent2(); //Uses the default constructor, so x=0 and y=0 (Not sure)
Entity ent3(1, 2); //Made constructor, so x=1 and y=2
Entity ent4 = Entity(); //Default constructor, so x=0 and y=0
Entity ent5 = Entity(2, 3); //Made constructor, so x=2 and y=3
我知道可以在堆内存上创建一个对象，但这不是我目前想要的.
I know that's it's possible to make an object on the heap memory, but that's not what I am looking for at this moment.
我的问题是，这些初始化对象的方式有什么区别?
My question is, what's the difference between these ways of initializing an object?
我不确定我应该在什么时候使用哪个.
I'm not sure which one I should use when.
推荐答案
Entity ent1;
上面的语句使用类Entity的默认构造函数.
The statement above uses default constructor of class Entity.
Entity ent2();
如果可能的话，上面的声明将被编译器视为函数原型.它被称为最令人烦恼的解析 (MVP) 的一个案例，它的存在导致出现误导性的聪明的愚蠢规则":永远不要使用括号".
The declaration above will be treated by compiler as a function prototype if that's possible. It's known as a case of most vexing parse (MVP) and its existence led to appearance of misleading "clever dumb rule": "never use parenthesis".
在这样的语句中，为 ent3 调用了一个用户定义的构造函数:
In statement like this a user-defined constructor is invoked for ent3:
Entity ent3(1, 2);
MVP 可以罢工的另一种情况是这样的:
Another case where MVP can strike is something like this:
Entity ent3_1(int(a), int(b)); // It's not what it looks like.
上面的
ent3_1 不是变量.该语句声明了一个带有两个 int 参数的函数.int(a) 与 int a 相同是 C 语言和声明语法的遗产.
ent3_1 above is not a variable. The statement declares a function with two int parameters. int(a) being same as int a is legacy of C language and declaration syntax there.
Entity ent4 = Entity();
ent4 是 C++11 之前 ent2 案例的正确 版本.默认构造函数作为值初始化的一部分被调用.它的形式允许避免使ent2和ent3_1不正确的歧义解决原则.这里的等号不是赋值，因为这里不会发生 operator= 调用.它是用于标记初始化表达式的声明语法的一部分.
ent4 is a proper version of ent2 case until C++11. Default constructor is invoked as part of value initialization. Its form allows to avoid an ambiguity solving principle which makes ent2 and ent3_1 incorrect. Equal sign here is not an assignment, for no operator= call will happen here. It's part of declaration syntax meant to markup the initialization expression.
Entity ent5 = Entity(2, 3);
ent5 是 ent3 case 的一个版本.作为值初始化的一部分调用的用户定义的构造函数.
ent5 is a version of ent3 case. User-defined constructor invoked as part of value initialization.
您的问题被标记为 C++11，而 C++11 允许统一初始化语法:
Your question is tagged as C++11, and C++11 allows uniform initialization syntax:
Entity ent12{}; // This is a legal alternative of ent2 case
Entity ent13{1, 2}; // A call to constructor or member initialization
Entity ent13{ int(a), int(b) }; // Not a function anymore
Entity ent14 = {}; // Not an assignment
Entity ent15 = Entity{2, 3}; // Not an assignment either!
请注意，统一初始化语法有一个警告.例如.这一行
Note that uniform initialization syntax has a caveat. E.g. this line
std::vector<int> v(10);
声明一个包含 10 个元素的向量.但是这个
declares a vector of 10 elements. But this one
std::vector<int> v{10};
声明一个向量，该向量使用值为 10 的 int 类型的单个元素进行初始化.发生这种情况是因为 std::vector 有一个定义了以下签名的构造函数:
declares a vector initialized with single element of type int with value 10. This happens because std::vector has a constructor with following signature defined:
vector( std::initializer_list<T> init, const Allocator& alloc = Allocator() );
如果您既不能在不触发 MVP 的情况下使用 () 也不能在不调用不需要的构造函数的情况下使用 {} 时，值初始化赋值语法可以解决该问题.
In case that you can't use neither () without triggering MVP nor {} without invoking undesired constructor, the value initialization assignment syntax allows to resolve the issue.
附录:必须观看 CppCon 2018:Nicolai JosuttisC++ 初始化的噩梦""
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