多维数组指针在 C 中的工作原理 [英] How Pointer of Multi-Dimension Arrays Work in C

问题描述

我正在试验 C 中指向多维​​数组的指针的概念.假设我想通过一个函数处理一个多维数组.代码有点像这样:

I'm experimenting with the concept of pointer to multi-dimension array in C. Suppose I want to process a multi-dimensional array via a function. The code kinda looks like this:

#include <stdio.h>

void proc_arr(int ***array)
{
    // some code
}

int main(int argc, char **argv)
{
    int array[10][10];
    for(int i = 0; i < 10; i++)
    {
        for(int j = 0; j < 10; j++)
        {
            array[i][j] = i * j;
        }
    }

    proc_arr(&array);

    return 0;
}

问题是，当我想访问 proc_arr 中的 array 时，我不能.根据我的理解，我们应该这样访问它:

The problem is, when I want to access array inside proc_arr, i can't. From my understanding, we should access it this way:

void proc_arr(int ***array)
{
    (*array)[0][1] = 10;
}

所以我取消引用 array 来告诉编译器我想要去那个地址并获取值.但不知何故，它崩溃了.我已经尝试了几种 * 和括号的组合，但仍然无法使其工作.我很确定这是因为我不理解指针和指针的指针.

So I derefer the array to tell the compiler that I want to go to that address and get the value. But somehow, it crashes. I've tried several combinations of * and parentheses and still can't make it work. I'm pretty sure it's because of me not understanding pointers and pointers of pointers.

哦，我注意到如果我们使用 char **(字符串数组)也是不同的，比如 argv 和 envp.但是对于 envp，我可以通过 (*envp) 以某种方式访问​​它.为什么?

Oh, and I've noticed that it's different if we work with a char **(array of string) too, like for argv and envp. But for envp, I somehow can access it with (*envp). Why?

这是 procces envp(并有效)的函数:

Here's the function that procces envp (and worked):

int envplen(char ***envp)
{
    int count = 0;

    while((*envp)[count] != NULL)
    {
        count++;
    }

    return count;
}

另外，我能否以某种方式访问​​ envplen 函数中的 envp 仅使用 envp，但仍然通过引用传递它?

Also, can I somehow access envp in the envplen function with only envp, but still pass it by reference?

之前谢谢.

推荐答案

问题是因为 int array[10][10] 分配在堆栈上并没有按照你想象的方式布置内存做.这是因为数组不是指针.内存仍然以线性阵列布局，而不是二维"阵列，即使下标可能表示如此.换句话说，int array[10][10] 的内存如下所示:

The problem is because int array[10][10] allocated on the stack does not lay out memory the way you think it does. This is because arrays are not pointers. The memory is still laid out in a linear array, not a "two dimensional" array, even though that's what the subscripts might indicate. In other words, the memory for int array[10][10] looks like the following:

starting address:                                    ending address:
| Block_1 of 10 int | Block_2 of 10 int | ... | Block_10 of 10 int |

因此，当您将数组隐式转换为 int***，然后尝试像 (*array)[1][10] 一样访问该数组时，这实际上意味着什么像*(*((*array) + 1) + 10)，这种操作的内存布局希望看到内存设置如下:

So when you implicitly convert the array to an int***, and then try to access the array like (*array)[1][10], what this actually translates to is something like *(*((*array) + 1) + 10), and the memory layout for such an operation wants to see memory setup like the following:

int*** array
|
|
| Pointer |
|
|
| Pointer_0 | Pointer_1 | ... | Pointer 10 |
       |          |                 |
       |          |                 | Block of 10 int |
       |          |
       |          | Block of 10 int |
       |
       |Block of 10 int|

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