C 中的 malloc，但使用多维数组语法 [英] malloc in C, but use multi-dimensional array syntax

问题描述

有什么办法可以 malloc 一个大数组，但是用 2D 语法引用它?我想要类似的东西:

Is there any way to malloc a large array, but refer to it with 2D syntax? I want something like:

int *memory = (int *)malloc(sizeof(int)*400*200);
int MAGICVAR = ...;
MAGICVAR[20][10] = 3; //sets the (200*20 + 10)th element

<小时>更新:这一点很重要:我只想拥有一个连续的内存块.我只是不想写一个像这样的宏:

---

UPDATE: This was important to mention: I just want to have one contiguous block of memory. I just don't want to write a macro like:

#define INDX(a,b) (a*200+b);

然后像这样引用我的 blob:

and then refer to my blob like:

memory[INDX(a,b)];

我更喜欢:

memory[a][b];

<小时>更新:我知道编译器无法按原样知道.我愿意提供额外的信息，例如:

---

UPDATE: I understand the compiler has no way of knowing as-is. I'd be willing to supply extra information, something like:

int *MAGICVAR[][200] = memory;

不存在这样的语法吗?请注意，我不只使用固定宽度数组的原因是它太大而无法放在堆栈中.<小时>更新:好的伙计们，我可以这样做:

Does no syntax like this exist? Note the reason I don't just use a fixed width array is that it is too big to place on the stack.

---

UPDATE: OK guys, I can do this:

void toldyou(char MAGICVAR[][286][5]) {
  //use MAGICVAR
}

//from another function:
  char *memory = (char *)malloc(sizeof(char)*1820*286*5);
  fool(memory);

我收到一个警告，从不兼容的指针类型传递了告诉你的 arg 1，但代码有效，我已经验证访问了相同的位置.有没有办法在不使用其他功能的情况下做到这一点?

I get a warning, passing arg 1 of toldyou from incompatible pointer type, but the code works, and I've verified that the same locations are accessed. Is there any way to do this without using another function?

推荐答案

是的，你可以这样做，不，你不需要像大多数其他答案告诉你的那样另一个指针数组.您想要的调用只是:

Yes, you can do this, and no, you don't need another array of pointers like most of the other answers are telling you. The invocation you want is just:

int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
MAGICVAR[20][10] = 3; // sets the (200*20 + 10)th element

<小时>

如果你想声明一个返回这样一个指针的函数，你可以这样做:

---

If you wish to declare a function returning such a pointer, you can either do it like this:

int (*func(void))[200]
{
    int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
    MAGICVAR[20][10] = 3;

    return MAGICVAR;
}

或者使用 typedef，这样会更清楚一点:

Or use a typedef, which makes it a bit clearer:

typedef int (*arrayptr)[200];

arrayptr function(void)
{
    /* ... */

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