存储器映射在C二维数组 [英] Memory map for a 2D array in C

问题描述

你觉得什么这个讨论了解二维数组的内存映射是正确的？ 尤其是照片你能解释一下原理是什么？

Do you think what this discussion about memory-map of 2D array is correct? Especially this photo? Can you explain the theory?

假设我们用C声明一个二维数组是这样的：

Suppose we declare a 2D array in C like this:

int arr[3][3]={10, 20, 30, 40, 50, 60, 70, 80, 90};

现在，根据本讨论中，存储器会安排如下所示：

Now, according to this discussion, the memory would be arranged like the following:

现在，我写了下面的code测试这种理论：

Now, I have written the following code to test this theory:

#include <stdio.h>

main()
{
    int arr[3][3]={10, 20, 30, 40, 50, 60, 70, 80, 90};
    printf("      arr==%d\n", arr);
    printf("  &arr[0]==%d\n", &arr[0]);
    printf("   arr[0]==%d\n", arr[0]);
    printf("&arr[0][0]=%d\n", &arr[0][0]);
    printf(" arr[0][0]=%d\n", arr[0][0]);
}
/*
Output:
========
      arr ==1245028
  &arr[0] ==1245028
   arr[0] ==1245028
&arr[0][0]==1245028
 arr[0][0]==10
Press any key to continue...
*/

为什么第4个输出相同？

Why the first 4 outputs are same?

推荐答案

您code只使用一个简单的多维数组，但图像描述指针数组，喜欢那种你通常当的malloc-ING的事情做。

Your code just uses a plain multidimensional array, but the image describes an array of pointers, like the kind you usually make when malloc-ing things.

一个多维数组基本上是一个正常的，扁平的，阵列（在内存中），以访问一些额外的语法糖。因此，尽管有可能从ARR [I]得到一个指针，没有一个额外的变量只是存储这一点，因为在你的形象发生了。

A multidimensional array is basically just a normal, flattened, array (in memory) with some extra syntatic sugar for accessing. So while it is possible to get a pointer from arr[i], there isn't an extra "variable" just to store this, as happens in your image.

要校正图像，用改编[0]，ARR [1] ... 和改变的值改编去除部分到 1245039 （同​​为＆放大器;常用3 [0] [0]）。

To correct the image, remove the parts with the arr[0], arr[1]... and change the value of arr to 1245039 (the same as &arr[0][0]).

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