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。

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。

你甚至可以把Gui13的解决方案打包成一个初始化语句:

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


address ta = (ta = address(), ta.city = "Hamilton", ta.prov = "Ontario", ta);

免责声明:我不推荐这种风格

如果你想清楚每个初始化值是什么，只需要把它分成多行，并在每一行上加一个注释:

address temp_addres = {
  0,  // street_no
  nullptr,  // street_name
  "Hamilton",  // city
  "Ontario",  // prov
  nullptr,  // postal_code
};

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);
}

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 };