尝试在 C 中创建数组的数组

发布于 2025-01-07 17:16:58 字数 498 浏览 0 评论 0原文

所以我有一堆布尔数组，我想将它们放入一个数组中以便于访问，但由于某种原因这不太有效。

我的数组如下所示：

boolean l1_000[8] = {1,0,0,0,0,0,0,0};

我声明我的数组数组：

boolean level1[8];

然后我想我这两个中的任何一个都可以（第一个只是直接声明这些数组，我将它们设置在大数组上）：

level1[0] = {1,0,0,0,0,0,0,0};
level1[0] = l1_000;

我也尝试了 level1[8][]，但这也不起作用。那么我在这里做错了什么？我该怎么做？

编辑：所以我设法通过将数组声明为 boolean *level1[8] 来做到这一点，但这只允许我执行 level1[0] = l1_000。有什么办法可以做到 level1[0] = {1,0,0,0,0,0,0,0} 吗？

原文

So I have a bunch of boolean arrays that I would like to put into a single array for easier accessing, but for some reason this doesn't quite work.

My arrays look like this:

boolean l1_000[8] = {1,0,0,0,0,0,0,0};

I declare my array of arrays with:

boolean level1[8];

And then I figured I could either of these two (first of which just declaring these arrays directly where I set them on the big array):

level1[0] = {1,0,0,0,0,0,0,0};
level1[0] = l1_000;

I also tried level1[8][], but that didn't work either. So what am I doing wrong here? How would I do this?

EDIT: So I managed to do this by declaring the array as boolean *level1[8], but that only allows me to do level1[0] = l1_000. Is there any way I can do level1[0] = {1,0,0,0,0,0,0,0}?

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空心空情空意 2025-01-14 17:16:58

当您想要存储已创建的数组而不复制每个元素时，不能声明数组的数组，但可以声明指针数组：

boolean* level1[] = {
    l1_000, // the array name decays to a pointer to the first element
    l2_000,
    // etc
};

You can't declare an array of arrays when you want to store arrays that are already created without copying each element, but you can declare an array of pointers:

boolean* level1[] = {
    l1_000, // the array name decays to a pointer to the first element
    l2_000,
    // etc
};

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下壹個目標 2025-01-14 17:16:58

我的 C-fu 有点弱，但你必须这样做：

int l1_000[8] = {1,0,0,0,0,0,0,0}; // one row
int level1[8][8];// 8 high, 8 wide
memcpy(level1[0], l1_000, sizeof(int)*8); // size of int * number of memory elements

基本上只有在你初始化它时才应该使用 int variable[] ，否则它不会工作。

像这样设置数组位置也

level1[0] = l1_000;

不起作用，因为您试图将第一个元素设置为数组。

My C-fu is kinda weak, but you have to do this:

int l1_000[8] = {1,0,0,0,0,0,0,0}; // one row
int level1[8][8];// 8 high, 8 wide
memcpy(level1[0], l1_000, sizeof(int)*8); // size of int * number of memory elements

basically doing int variable[] will should only be used when you are initializing it right then and there other wise it won't work.

also setting the array position like this