c++ - 带大括号的变量初始化

Question

是以下语法:

T t = {};

对于非 POD 类型是零初始化的零初始化还是值初始化？

Answer 1

语法 T object = {arg1, arg2, ...};是 copy list initialization .自从您建立了T不是聚合类型，并且 T有一个默认构造函数(来自列表)然后列表初始化将执行 value initialization :

• If T is an aggregate type, aggregate initialization is performed.
• Otherwise, if the braced-init-list is empty and T is a class type with a default constructor, value-initialization is performed.