我可能在这里遗漏了一些东西，为什么不呢:

#include <cstdio>    
struct Group {
    int x;
    int y;
    const char* s;
};

int main() 
{  
  Group group {
    .x = 1, 
    .y = 2, 
    .s = "Hello it works"
  };
  printf("%d, %d, %s", group.x, group.y, group.s);
}

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?

我知道这个问题很老了，但我找到了另一种初始化的方法，使用constexpr和currying:

struct mp_struct_t {
    public:
        constexpr mp_struct_t(int member1) : mp_struct_t(member1, 0, 0) {}
        constexpr mp_struct_t(int member1, int member2, int member3) : member1(member1), member2(member2), member3(member3) {}
        constexpr mp_struct_t another_member(int member) { return {member1, member, member3}; }
        constexpr mp_struct_t yet_another_one(int member) { return {member1, member2, member}; }

    int member1, member2, member3;
};

static mp_struct_t a_struct = mp_struct_t{1}
                           .another_member(2)
                           .yet_another_one(3);

此方法也适用于全局静态变量，甚至是构造变量。 唯一的缺点是糟糕的可维护性:每次必须使用此方法使另一个成员可初始化时，所有成员初始化方法都必须更改。

这个特性称为指定初始化式。它是C99标准的补充。然而，这个特性在c++ 11中被忽略了。根据c++编程语言，第4版，第44.3.3.2节(c++未采用的C特性):

c++故意没有采用C99的一些附加功能(与C89相比): [1]变长数组(VLAs);使用矢量或某种形式的动态数组 [2]指定初始化式;使用构造函数

C99语法有指定的初始化式[参见ISO/IEC 9899:2011, N1570委员会草案- 2011年4月12日]

6.7.9初始化

initializer:
    assignment-expression
    { initializer-list }
    { initializer-list , }
initializer-list:
    designation_opt initializer
    initializer-list , designationopt initializer
designation:
    designator-list =
designator-list:
    designator
    designator-list designator
designator:
    [ constant-expression ]
    . identifier

另一方面，c++ 11没有指定的初始化式[参见ISO/IEC 14882:2011, N3690委员会草案- 2013年5月15日]

8.5初始化

initializer:
    brace-or-equal-initializer
    ( expression-list )
brace-or-equal-initializer:
    = initializer-clause
    braced-init-list
initializer-clause:
    assignment-expression
    braced-init-list
initializer-list:
    initializer-clause ...opt
    initializer-list , initializer-clause ...opt
braced-init-list:
    { initializer-list ,opt }
    { }

为了达到同样的效果，可以使用构造函数或初始化列表:

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。

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。)