class 聚合的元素
Elements of a class aggregate
我试图了解从 C++14 到 C++17 发生的关于聚合初始化的变化。来自 §9.4.2[dcl.init.aggr]/2:
The elements of an aggregate are:
- (2.1) for an array, the array
elements in increasing subscript order, or
- (2.2) for a class, the
direct base classes in declaration order, followed by the direct
non-static data members ([class.mem]) that are not members of an
anonymous union, in declaration order.
本段定义聚合 class 的元素是什么。特别是，我无法理解粗体句子的含义。我有这个例子来证明我的困惑：
struct Base { int mem1, mem2; };
struct Derived : Base { int mem3, mem4; };
Derived obj = { 1, 2, 3, 4 };
对象 obj 是一个集合，因为它满足 [dcl.init.aggr]/1 中定义的条件。然而，我的问题是这个聚合的元素是什么。好像元素是Base::mem1初始化为1，Base::mem2初始化为2，Derived::mem3初始化为3，Dervied::mem4初始化为4.
但是 [dcl.init.aggr]/2 的说法恰恰相反。据说元素是直接基数classes的数据成员，在本例中是Base，直接非静态数据后跟[=33=] Derived class 的成员。我是否正确解析了措辞？
我真的不知道我是否正确理解了这句话，但我想我没有。所以关于这一点的任何解释都会对我有很大帮助。
您对“聚合元素”概念的理解是正确的。然而，这并不是聚合初始化的结束。有brace elision:
Braces can be elided in an initializer-list as follows. If the initializer-list begins with a left brace, then the succeeding comma-separated list of initializer-clauses initializes the elements of a subaggregate; it is erroneous for there to be more initializer-clauses than elements. If, however, the initializer-list for a subaggregate does not begin with a left brace, then only enough initializer-clauses from the list are taken to initialize the elements of the subaggregate; any remaining initializer-clauses are left to initialize the next element of the aggregate of which the current subaggregate is an element.
All implicit type conversions ([conv]) are considered when initializing the element with an assignment-expression. If the assignment-expression can initialize an element, the element is initialized. Otherwise, if the element is itself a subaggregate, brace elision is assumed and the assignment-expression is considered for the initialization of the first element of the subaggregate.
由于 Base b = 1; 不是有效的初始化语句，因此假定大括号省略，如第一段中所定义。
我试图了解从 C++14 到 C++17 发生的关于聚合初始化的变化。来自 §9.4.2[dcl.init.aggr]/2:
The elements of an aggregate are:
- (2.1) for an array, the array elements in increasing subscript order, or
- (2.2) for a class, the direct base classes in declaration order, followed by the direct non-static data members ([class.mem]) that are not members of an anonymous union, in declaration order.
本段定义聚合 class 的元素是什么。特别是，我无法理解粗体句子的含义。我有这个例子来证明我的困惑：
struct Base { int mem1, mem2; };
struct Derived : Base { int mem3, mem4; };
Derived obj = { 1, 2, 3, 4 };
对象 obj 是一个集合，因为它满足 [dcl.init.aggr]/1 中定义的条件。然而，我的问题是这个聚合的元素是什么。好像元素是Base::mem1初始化为1，Base::mem2初始化为2，Derived::mem3初始化为3，Dervied::mem4初始化为4.
但是 [dcl.init.aggr]/2 的说法恰恰相反。据说元素是直接基数classes的数据成员，在本例中是Base，直接非静态数据后跟[=33=] Derived class 的成员。我是否正确解析了措辞？
我真的不知道我是否正确理解了这句话，但我想我没有。所以关于这一点的任何解释都会对我有很大帮助。
您对“聚合元素”概念的理解是正确的。然而，这并不是聚合初始化的结束。有brace elision:
Braces can be elided in an initializer-list as follows. If the initializer-list begins with a left brace, then the succeeding comma-separated list of initializer-clauses initializes the elements of a subaggregate; it is erroneous for there to be more initializer-clauses than elements. If, however, the initializer-list for a subaggregate does not begin with a left brace, then only enough initializer-clauses from the list are taken to initialize the elements of the subaggregate; any remaining initializer-clauses are left to initialize the next element of the aggregate of which the current subaggregate is an element.
All implicit type conversions ([conv]) are considered when initializing the element with an assignment-expression. If the assignment-expression can initialize an element, the element is initialized. Otherwise, if the element is itself a subaggregate, brace elision is assumed and the assignment-expression is considered for the initialization of the first element of the subaggregate.
由于 Base b = 1; 不是有效的初始化语句，因此假定大括号省略，如第一段中所定义。