是否可以在c++中初始化结构,如下所示:

struct address {
    int street_no;
    char *street_name;
    char *city;
    char *prov;
    char *postal_code;
};

address temp_address = { .city = "Hamilton", .prov = "Ontario" };

这里和这里的链接提到,这种样式只能在C中使用。如果是这样,为什么在c++中不能使用呢?是否有任何潜在的技术原因,为什么它不是在c++中实现的,或者使用这种风格是不好的做法。我喜欢使用这种初始化方式,因为我的结构体很大,而且这种样式可以让我清楚地了解分配给哪个成员的值。

请与我分享是否有其他方法可以达到同样的可读性。

在提出这个问题之前,我已参考以下连结:

C/ c++ for AIX C结构初始化变量 c++中使用标记的静态结构初始化 c++ 11正确的结构初始化


当前回答

它不是在c++中实现的。(同样,char* strings?我希望不会)。

通常情况下,如果你有这么多参数,那就是相当严重的代码味道。但是,为什么不简单地对结构进行值初始化,然后为每个成员赋值呢?

其他回答

它不是在c++中实现的。(同样,char* strings?我希望不会)。

通常情况下,如果你有这么多参数,那就是相当严重的代码味道。但是,为什么不简单地对结构进行值初始化,然后为每个成员赋值呢?

这是可能的,但前提是初始化的结构体是POD(普通旧数据)结构体。它不能包含任何方法、构造函数,甚至默认值。

我发现这种方式做全局变量,不需要修改原来的结构定义:

struct address {
             int street_no;
             char *street_name;
             char *city;
             char *prov;
             char *postal_code;
           };

然后声明一个继承自原始结构类型的新类型变量,并使用构造函数初始化字段:

struct temp_address : address { temp_address() { 
    city = "Hamilton"; 
    prov = "Ontario"; 
} } temp_address;

虽然没有C风格那么优雅…

对于局部变量,它需要在构造函数的开头添加一个额外的memset(this, 0, sizeof(*this)),所以它显然并不差,@gui13的答案更合适。

(注意,'temp_address'是一个'temp_address'类型的变量,但是这个新类型继承自'address',可以在任何需要'address'的地方使用,所以它是OK的。)

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。

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 };