结构变量成员后的大括号是什么意思？
What do curly braces after a struct variable member mean?
在一些维护（Valgrind'ing）期间，我遇到了这段代码：
#pragma pack(push, 1)
struct somename
{
uint16_t a{};
uint16_t b{};
uint32_t c{};
};
#pragma pack(pop)
我希望 {} 告诉编译器始终将值初始化为 0（当使用 new 进行分配或使用堆栈变量时），但我找不到任何相关示例或文档。我的这个假设是否正确？如果不是：
struct成员变量后面的大括号{}是什么意思？
这是 default member initializer（C++11 起）。
（强调我的）
Through a default member initializer, which is a brace or equals initializer included in the member declaration and is used if the member is omitted from the member initializer list of a constructor.
If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored for that constructor.
因此，数据成员 a、b 和 c 对于内置类型是 value-initialized (zero-initialized）到 0。
如文档所述，它是零初始化here（第二种情况）：
因此所有值都设置为 0。
来自docs：
Value initialization is performed when a named variable (automatic, static, or thread-local) is declared with the initializer consisting of a pair of braces T{}.
值初始化的效果是对象被零初始化。
在一些维护（Valgrind'ing）期间，我遇到了这段代码：
#pragma pack(push, 1)
struct somename
{
uint16_t a{};
uint16_t b{};
uint32_t c{};
};
#pragma pack(pop)
我希望 {} 告诉编译器始终将值初始化为 0（当使用 new 进行分配或使用堆栈变量时），但我找不到任何相关示例或文档。我的这个假设是否正确？如果不是：
struct成员变量后面的大括号{}是什么意思？
这是 default member initializer（C++11 起）。
（强调我的）
Through a default member initializer, which is a brace or equals initializer included in the member declaration and is used if the member is omitted from the member initializer list of a constructor.
If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored for that constructor.
因此，数据成员 a、b 和 c 对于内置类型是 value-initialized (zero-initialized）到 0。
如文档所述，它是零初始化here（第二种情况）：
因此所有值都设置为 0。
来自docs：
Value initialization is performed when a named variable (automatic, static, or thread-local) is declared with the initializer consisting of a pair of braces
T{}.
值初始化的效果是对象被零初始化。