问题描述
我有一个关于C / C ++如何在内部存储多维数组使用标记富[M]宣布的问题[N] 。我并不是质疑纯指针的指针等...我问的，因为速度的原因...
I have a question about how C / C++ internally stores multidimensional arrays declared using the notation foo[m][n]. I am not questioning pure pointers to pointers etc... I am asking because of speed reasons...
纠正我，如果我错了，但是从语法富是一个指针数组，这本身指向数组
Correct me if I am wrong, but syntactically foo is an array of pointers, which themselves point to an array
int foo[5][4]
*(foo + i) // returns a memory address
*( *(foo + i) + j) // returns an int
我已经从很多地方听到的C / C ++编译器转换富[M] [N] 来在幕后的一维数组（计算所需的一维与索引i *宽+ J ）。但是，如果这是真的，那么下面将举行
I have heard from many places that the C/C++ compiler converts foo[m][n] to a one dimensional array behind the scenes (calculating the required one dimension index with i * width + j). However if this was true then the following would hold
*(foo + 1) // should return element foo[0][1]
因此，我的问题：
这是真的，富[M] [N] 是（总是？）存储在内存中的一台一维数组？如果是这样，为什么如上述code的工作。
Thus my question:Is it true that foo[m][n] is (always?) stored in memory as a flat one dimensional array?? If so, why does the above code work as shown.
推荐答案
是的，C / C ++存储多维（矩形）阵列作为一个连续的内存区域。但是，你的语法不正确。要修改元素 foo的[0] [1] ，以下code将工作：
Yes, C/C++ stores a multi-dimensional (rectangular) array as a contiguous memory area. But, your syntax is incorrect. To modify element foo[0][1], the following code will work:
*((int *)foo+1)=5;
显式类型转换是必要的，因为富+ 1 ，是一样的＆放大器;富[1] 这是不是在所有一回事 foo的[0] [1] 。 *（美孚+ 1）是一个指向平面内存区第五元素。换句话说， *（美孚+ 1）基本 foo的[1] 和 * *（美孚+ 1）是 foo的[1] [0] 。下面是内存是如何为你的一些二维阵列的布局：
The explicit cast is necessary, because foo+1, is the same as &foo[1] which is not at all the same thing as foo[0][1]. *(foo+1) is a pointer to the fifth element in the flat memory area. In other words, *(foo+1) is basically foo[1] and **(foo+1) is foo[1][0]. Here is how the memory is laid out for some of your two dimensional array:
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