正如其他人提到的，这是指定初始化项。

这个特性是c++ 20的一部分

在c++中，C风格的初始化式被构造函数所取代，构造函数在编译时可以确保只执行有效的初始化(即初始化后对象成员是一致的)。

这是一个很好的实践，但有时预初始化也很方便，就像在您的示例中一样。OOP通过抽象类或创建设计模式解决了这个问题。

在我看来，使用这种安全的方式消除了简单性，有时安全性的权衡可能过于昂贵，因为简单的代码不需要复杂的设计来保持可维护性。

作为另一种解决方案，我建议使用lambdas来定义宏，以简化初始化，使其看起来几乎像c风格:

struct address {
  int street_no;
  const char *street_name;
  const char *city;
  const char *prov;
  const char *postal_code;
};
#define ADDRESS_OPEN [] { address _={};
#define ADDRESS_CLOSE ; return _; }()
#define ADDRESS(x) ADDRESS_OPEN x ADDRESS_CLOSE

ADDRESS宏展开为

[] { address _={}; /* definition... */ ; return _; }()

它创建并调用lambda。宏参数也是用逗号分隔的，因此需要将初始化式放在括号中并调用like

address temp_address = ADDRESS(( _.city = "Hamilton", _.prov = "Ontario" ));

你也可以写广义宏初始化式

#define INIT_OPEN(type) [] { type _={};
#define INIT_CLOSE ; return _; }()
#define INIT(type,x) INIT_OPEN(type) x INIT_CLOSE

但这样的呼唤就不那么美妙了

address temp_address = INIT(address,( _.city = "Hamilton", _.prov = "Ontario" ));

但是你可以很容易地使用INIT宏定义ADDRESS宏

#define ADDRESS(x) INIT(address,x)

受到这个非常简洁的答案的启发:(https://stackoverflow.com/a/49572324/4808079)

你可以使用lamba闭包:

// Nobody wants to remember the order of these things
struct SomeBigStruct {
  int min = 1;
  int mean = 3 ;
  int mode = 5;
  int max = 10;
  string name;
  string nickname;
  ... // the list goes on
}

.

class SomeClass {
  static const inline SomeBigStruct voiceAmps = []{
    ModulationTarget $ {};
    $.min = 0;  
    $.nickname = "Bobby";
    $.bloodtype = "O-";
    return $;
  }();
}

或者，如果你想要非常花哨的话

#define DesignatedInit(T, ...)\
  []{ T ${}; __VA_ARGS__; return $; }()

class SomeClass {
  static const inline SomeBigStruct voiceAmps = DesignatedInit(
    ModulationTarget,
    $.min = 0,
    $.nickname = "Bobby",
    $.bloodtype = "O-",
  );
}

这样做有一些缺点，主要与未初始化的成员有关。从链接的答案评论说，它编译有效，虽然我没有测试它。

总的来说，我认为这是一个很好的方法。

字段标识符实际上是C初始化式语法。在c++中，只需要按正确的顺序给出值，而不需要字段名。不幸的是，这意味着你需要给他们所有(实际上你可以省略后面的零值字段，结果将是相同的):

address temp_address = { 0, 0, "Hamilton", "Ontario", 0 }; 

正如其他人提到的，这是指定初始化项。

这个特性是c++ 20的一部分

你有

The standard initialization list address temp_address { /* street_no */, /* street_name */, ... /* postal_code */ }; address temp_address2 = { /* street_no */, /* street_name */, ... /* postal_code */ } The dot notation address temp_address; temp_address.street_no = ...; temp_address.street_name = ...; ... temp_address.postal_code = ...; The designated aggregate initialization, where the initialization list contains that labels of each member of the structure (see documentation) available from C++20 onward. Treating a struct like a C++ class - in C++ structures are actually special types of classes, where all members are public (unlike a standard C++ class where all members are private if not specified otherwise explicitly) as well as that when using inheritance they default to public: struct Address { int street_no; ... char* postal_code; Address (int _street_no, ... , char* _postal_code) : street_no(_street_no), ... postal_code(_postal_code) {} } ... Address temp_address ( /* street_no */, ..., /* postal_code */);

当涉及到初始化结构的方式时，你应该考虑以下方面:

Portability - different compilers, different degree of C++ standard completeness and different C++ standards altogether do limit your options. If you have to work with let's say a C++11 compiler but want to use the C++20 designated aggregate initialization you are out of luck Readability - what is more readable: temp_address.city = "Toronto" or temp_address { ..., "Toronto", ... }? Readability of your code is very important. Especially when you have large structures (worse - nested ones), having unlabeled values all over the place is just asking for trouble Scalability - anything that depends on a specific order is not a good idea. The same goes for lack of labels. You want to move a member up or down the address space of the structure? Good luck with an unlabeled initialization list (hunting down swapped values in structure initialization is a nightmare)... You want to add a new member? Again good luck with anything that depends on a specific order.

虽然点表示法意味着你输入更多，但你从使用它中得到的好处超过了这个问题，因此我建议你使用它，除非你有一个小的结构，它的结构缺乏变化，在这种情况下，你可以使用一个初始化列表。记住:无论何时与他人合作，编写易于遵循的代码都是至关重要的。