C中静态变量的初始化 [英] The initialization of static variables in C

问题描述

我有一个关于 C 中静态变量初始化的问题.我知道如果我们声明一个全局静态变量，默认值是 0.例如:

I have a question about the initialization of static variables in C. I know if we declare a global static variable that by default the value is 0. For example:

static int a; //although we do not initialize it, the value of a is 0

但是下面的数据结构呢:

but what about the following data structure:

typedef struct
{
    int a;
    int b;
    int c;
} Hello;

static Hello hello[3];

是hello[0]、hello[1]、hello[2]的每个结构体中的所有成员初始化为<代码>0?

are all of the members in each struct of hello[0], hello[1], hello[2] initialized as 0?

推荐答案

是的，所有成员都针对具有静态存储的对象进行了初始化.参见 C99 标准(PDF 文档)中的 6.7.8/10)

Yes, all members are initialized for objects with static storage. See 6.7.8/10 in the C99 Standard (PDF document)

如果一个具有自动存储持续时间的对象没有显式初始化，它的值是不确定的.如果没有显式初始化具有静态存储持续时间的对象，则:
——如果是指针类型，则初始化为空指针；
— 如果是算术类型，则初始化为(正或无符号)零；
— 如果是聚合，则根据这些规则(递归地)初始化每个成员；
——如果是联合，第一个命名成员根据这些(递归)初始化规则.

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then:
— if it has pointer type, it is initialized to a null pointer;
— if it has arithmetic type, it is initialized to (positive or unsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules;
— if it is a union, the first named member is initialized (recursively) according to these rules.

要将对象中的所有内容(无论是否为 static)初始化为 0，我喜欢使用 通用零初始化器

To initialize everything in an object, whether it's static or not, to 0, I like to use the universal zero initializer

sometype identifier0 = {0};
someothertype identifier1[SOMESIZE] = {0};
anytype identifier2[SIZE1][SIZE2][SIZE3] = {0};

<小时>

C 中没有部分初始化.一个对象要么完全初始化(在没有不同值的情况下为正确类型的0)，要么没有在全部.
如果要部分初始化，就不能初始化.

---

There is no partial initialization in C. An object either is fully initialized (to 0 of the right kind in the absence of a different value) or not initialized at all.
If you want partial initialization, you can't initialize to begin with.

int a[2]; // uninitialized
int b[2] = {42}; // b[0] == 42; b[1] == 0;
a[0] = -1; // reading a[1] invokes UB

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