c++ - 字符串和字符串数组的零初始化(C++)

Question

根据 https://en.cppreference.com/w/cpp/language/zero_initialization

在文档提供的例子中:

std::string s; // is first zero-initialized to indeterminate value
               // then default-initialized to ""

如果语法用于static T object;，为什么string s; 会出现零初始化？

为什么零初始化发生在默认初始化之前，为什么两者都允许发生？

零初始化的效果是:

• If T is a scalar type, the object's initial value is the integral constant zero explicitly converted to T.
• If T is an non-union class type, all base classes and non-static data members are zero-initialized, and all padding is initialized to zero bits. The constructors, if any, are ignored.
• If T is a union type, the first non-static named data member is zero-initialized and all padding is initialized to zero bits.
• If T is array type, each element is zero-initialized
• If T is reference type, nothing is done.

如果我初始化 string array[2] = {"Test1"}; 会怎样？我知道该数组将包含“Test1”和空字符串“”。

但是根据上面的文档，

If T is array type, each element is zero-initialized

数据类型是字符串，是对象/引用类型？

If T is reference type, nothing is done.

什么都没做？我想也许会调用一个构造函数。肯定是空字符串吧？

Answer 1

(除非另有说明，否则假定此答案中的所有声明都在命名空间范围内。)

Why does zero initialization occur to string s; if the syntax is for static T object;?
Why does zero initialization happen before default initialization and why are both allowed to happen?

具有静态存储持续时间的变量首先在编译时被零初始化，然后可选择地在运行时动态初始化。 static T object; 声明一个静态存储持续时间的对象。对于像

这样的简单声明

int x;

不执行动态初始化。对于更复杂的声明，例如

std::string s;

对字符串进行零初始化可能会导致类不变量损坏的无效字符串。因此，动态初始化调用默认构造函数来保证对象有效。

What if I initialize string array[2] = {"Test1"};? I know that the array will contain "Test1" and empty string "".

首先，在编译时，两个对象被零初始化，导致可能的无效状态。然后，在运行时调用构造函数(const char* 第一个对象的构造函数和第二个对象的默认构造函数)，并构造有效的对象。

The data type is string which is an object / reference type?

std::string 是对象类型而不是引用类型。

[For a reference type] Nothing is done? I thought maybe a constructor would have been called. Surely an empty string is something?

引用类型不被视为实际的“对象”，因此没有必要指定其零初始化语义。