如何多维数组工作中的C指针

本文介绍了如何多维数组工作中的C指针的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我与指针的概念进行实验，以C.多维数组假设我想处理通过功能的多维数组。在code还挺看起来是这样的：

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，当我要访问阵列在 proc_arr ，我不能。从我的理解，我们应该访问它是这样的：

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derefer的阵列来告诉我要到该地址并获得价值的编译器。但不知何故，它崩溃。我试过的 * 和括号几个组合仍然不能使它发挥作用。我是pretty确保它是因为我不理解指针的指针和指针。

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，这是不同的，如果我们有一个的char **工作（字符串数组）太像了的argv和envp。但对于envp，不知何故，我可以用访问（* envp）。为什么呢？

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?

下面是procces envp的函数（工作）：

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，我可以以某种方式访问​​ envp 在 envplen 函数仅 envp ，但还是按引用传递呢？

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

谢谢了。

推荐答案

现在的问题是，因为 int数组[10] [10] 在栈中分配并不铺陈上内存你觉得它的方式。这是因为的数组不是指针的。内存仍然是一个线性阵列布局，而不是一个二维阵列，即使这就是下标可能表示。换句话说，对于 int数组的存储[10] [10] 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 |

所以，当你的隐式数组转换为 INT *** ，然后尝试访问像（*数组）数组[1] [10]，是什么这实际上转化为是像 *（*（（*数组）+ 1）+ 10），并为这种操作内存布局希望看到内存设置，如以下内容：

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|