用C多维数组：他们是锯齿状？ [英] Multi-Dimensional Arrays in C: are they jagged?

问题描述

关于C编程语言的一个简单的问题（ANSI-C）：

在C多维数组的锯齿状？

我的意思是 - 我们谈论的（指向其他地址在内存中的一个数组）数组的数组，或者这仅仅是长一维数组（这是按顺序存储在内存中）<？ / p>

什么困扰我的是，我有点相信：

矩阵[i] [j]的等同于 *（*（矩阵+ I）+ J）

解决方案

在C A多维数组是连续的。以下内容：

  INT米[4] [5];
 

由4 INT [5] ■在内存布局彼此相邻。

指针数组：

 为int *米[4];
 

参差不齐。每个指针可以指向（第一元件）不同长度的一个单独的阵列

M [] [J] 等同于 *（*（M + 1）+ J） 。请参阅 C11标准，部分6.5.2.1：

  

下标运算符[]的德网络nition是E1 [E2]是相同的（*（（E1）+（E2）））

因此​​， M [] [J] 等同于（*（M + 1））[J] ，相当于 *（*（M + 1）+ J）。

这等同存在，因为在大多数情况下，阵列式衰减前pressions的指针他们的第一个元素（C11标准6.3.2.1）。 M [] [J] 是间preTED如下：

• M 是一个数组的数组，所以它衰变的指针 M [0] ，第一子阵。


• M + 1 是一个指向 I 次的子阵 M 。


• M [] 等同于 *（M + 1），取消引用一个指针 I 个 M 。由于这是数组类型的前pression，它衰变的指针 M [] [0] 。


• M [] [J] 等同于 *（*（M + 1）+ J） ，解引用指针Ĵ日的的元素i 次的子阵 M 。

需要注意的是指向数组的指针从指针到他们的第一个元素不同。 M + 1 是一个指向数组的指针;它不是数组类型的前pression，并且它不衰减，是否一指针的指针或任何其他类型的

A simple question about the C programming language (ANSI-C):

Are the multi-dimensional arrays in C jagged?

I mean - are we talking about "array of arrays" (one array of pointers to other addresses in the memory) , or this is just "long one-dimensional array" (which is stored sequentially in the memory)?

What that bothers me is that I'm kinda sure that:

matrix[i][j] is equivalent to * ( * (matrix + i) + j)

解决方案

A multidimensional array in C is contiguous. The following:

int m[4][5];

consists of 4 int[5]s laid out next to each other in memory.

An array of pointers:

int *m[4];

is jagged. Each pointer can point to (the first element of) a separate array of a different length.

m[i][j] is equivalent to *(*(m+i)+j). See the C11 standard, section 6.5.2.1:

The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2)))

Thus, m[i][j] is equivalent to (*(m+i))[j], which is equivalent to *(*(m+i)+j).

This equivalence exists because in most contexts, expressions of array type decay to pointers to their first element (C11 standard, 6.3.2.1). m[i][j] is interpreted as the following:

• m is an array of arrays, so it decays to a pointer to m[0], the first subarray.
• m+i is a pointer to the ith subarray of m.
• m[i] is equivalent to *(m+i), dereferencing a pointer to the ith subarray of m. Since this is an expression of array type, it decays to a pointer to m[i][0].
• m[i][j] is equivalent to *(*(m+i)+j), dereferencing a pointer to the jth element of the ith subarray of m.

Note that pointers to arrays are different from pointers to their first element. m+i is a pointer to an array; it is not an expression of array type, and it does not decay, whether to a pointer to a pointer or to any other type.

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