C 中的静态初始化

发布于 2024-08-20 23:51:45 字数 333 浏览 5 评论 0原文

我想要看起来像这样的代码...

static linked_list* globalListHoldingAllSortsOfGoodies = initialize_linked_list();

/* [In a different file...] */
static int placeholder = add_to_global_list(goodies);

但是在 C 中非常量初始化是不可能的。

是否有某种方法可以在不破坏 C89 的情况下获得相同的效果？

关键是通过使用也使用占位符的宏声明好东西，让不同的事物“自动注册”到全局列表中。

原文

I want to have code that looks something like this...

static linked_list* globalListHoldingAllSortsOfGoodies = initialize_linked_list();

/* [In a different file...] */
static int placeholder = add_to_global_list(goodies);

But non-constant initialization is impossible in C.

Is there some way to get the same affect without breaking C89?

The point is to have different things "automatically register" themselves into the global list by declaring the goodies with a macro that also uses placeholder.

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慢慢从新开始 2024-08-27 23:51:45

您可以从静态数据构建链接列表。在 ANSI C89（又名 ISO C90）中，它可能如下所示：

struct node
{
 int data;
 struct node *next;
};

struct node nodes[] = { { 42, &nodes[1] }, { 137, 0 } };
struct node *list = nodes;

在 ISO C99（2000 年由 ANSI 采用）中，您还可以使用复合文字，例如，将

struct node *list = &(struct node){ 42, &(struct node){ 137, 0 } };

静态分配的节点与动态分配的节点混合是有问题的，因为释放前者会导致在未定义的行为中，因此有必要跟踪哪些节点属于哪个组。

You can build a linked list from static data. In ANSI C89 (aka ISO C90), it could look like this:

struct node
{
 int data;
 struct node *next;
};

struct node nodes[] = { { 42, &nodes[1] }, { 137, 0 } };
struct node *list = nodes;

In ISO C99 (adopted by ANSI in 2000), you can additionally use compound literals, eg

struct node *list = &(struct node){ 42, &(struct node){ 137, 0 } };

Mixing statically allocated nodes with dynamically allocated nodes is problematic as freeing the former will result in undefined behaviour, so it will be necessary to keep track of which nodes belong to which group.

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