字符串和字符串数组的零初始化（C++）

Question

According to https://en.cppreference.com/w/cpp/language/zero_initialization

In the example provided by the documentation:

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

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?

The effects of zero initialization are:

• 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.

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

But according to the above documentation,

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

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

If T is reference type, nothing is done.

Nothing is done? I thought maybe a constructor would have been called. Surely an empty string is something?

Answer 1

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?

In page you linked to, that defines a non-local variable.

Non-local variables are initialized in two phases.

1. Static intialization.
2. Dynamic initialization, if it applies.

In static initialization phase, a variable is initialized using constant initialization or zero initialization

Dyanmic initialization is used, if it applies, such as for objects that have the appropriate constructor or for objects that are initialized using an expression that can be evaulated at run time.

You can read more on the topic at https://en.cppreference.com .

Nothing is done? I thought maybe a constructor would have been called. Surely an empty string is something?

A reference cannot be zero-initialized. It can only be initialized using a object that it will be a reference to.

Answer 2

(Unless otherwise specified, all declarations in this answer are assumed to be in namespace scope.)

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?

Variables with static storage duration are first zero-initialized at compile time, and then optionally dynamically initialized at runtime. static T object; declares an object of static storage duration. For a simple declaration like

int x;

The dynamic initialization is not performed. For a more sophisticated declaration like

std::string s;

Zero-initializing a string may result in an invalid string with a broken class invariant. Therefore, the dynamic initialization calls the default constructor to ensure that the object is valid.

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

First, at compile time, the two objects are zero-initialized, resulting in possible invalid state. Then, at runtime, the constructors are called ( const char* constructor for the first object and default constructor for the second object), and the valid objects are constructed.

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

std::string is an object type instead of a reference type.

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

A reference type is not considered an actual "object", so there is no point in specifying its zero-initialization semantics.