是什么INT之间的差* X [n]的[米]和INT（* X）[n]的[米] [英] What is the difference between int* x[n][m] and int (*x) [n][m]?

问题描述

在我看来 INT * X [N] [M] 声明 X 是一个2-D指向整数数组，所以分配内存应该是那么容易，因为 X [I] [J] =新INT ，并如预期正常工作。现在，如果我改变声明：

As I see it int *x[n][m] declares x to be a 2-d array of pointers to integers, so allocating memory should be as easy as x[i][j] = new int and as expected it works fine. Now if I change the declaration to:

INT（* X）[N] [M]

X [I] [J] =新INT 不再起作用并导致编译错误。

x[i][j] = new int no longer works and results in a compilation error.

X =（INT（*）[N] [M]）的malloc（sizeof的（INT [N] [M]））然而编译。从我已运行了几个测试，内存分配后，不同的报关/分配组合似乎并不影响存储在变量的值。我缺少的东西吗？所以我的问题是，有* X [N] [M]和INT（* X）[M] INT之间的差异[N]。如何为int（* X）[N] [M]存储在内存中？

x = (int(*)[n][m]) malloc (sizeof(int[n][m])) however compiles. From the few tests that I have run, after the memory allocation, the different declaration/allocation combination doesn't seem to effect the values stored in the variable. Am I missing something? So my question is, Is there a difference between int *x[n][m] and int (*x)[m][n]. How is int (*x)[n][m] stored in memory?

推荐答案

为int *一个[N] [M] 是指针的二维数组 INT 。

INT（* P）[N] [M] 是一个指向 INT 的二维数组S（这是你采取的地址获取类型的INT [n] [M] ）。

int (*p)[n][m] is a pointer to a two dimensional array of ints (it is the type you get by taking the address of int[n][m]).

在这两种情况下， N 和 M 必须编译时间常数，否则声明是不合法的C ++（但在C）。你的编译器可能有一个扩展，允许它，虽然。

In both cases, n and m need to be compile time constants, otherwise the declarations are not legal in C++ (but are in C). Your compiler might have an extension to allow it, though.

首先可以用来模拟一个三维阵列。我说模仿的，因为它不会与连续的存储适当的阵列和类型是摆在首位的不同。在每个 A 您可以存储地址到整数数组的第一个元素的元素。每个可以具有不同的尺寸和动态分配。您可以将存储指向一个单一的（可能是堆栈中分配）整了。

First one can be used to simulate a three dimensional array. I say simulate, because it would not be a proper array with contiguous storage and the types are different in the first place. In each of the elements of a you can store the address to the first element of an array of integers. Each could have a different size and be allocated dynamically. You can store a pointer to a single (possibly stack allocated) integer, too.

int i = 0;
int a1[2] = {};

int* a2[2][2];
a2[0][0] = a1;  // array to pointer decay here
a2[0][1] = new int[42];
a2[1][0] = new int[84];
a2[1][1] = &i;

P 可以指向一个二维数组或其数组：

p can point to a single 2d array or an array thereof:

int arr[2][3];
int (*p1)[2][3] = &arr;  // arr decays to int(*)[3], so we need to take the address
int (*p2)[2][3] = new int[42][2][3]; // allocate 42 arrays dynamically

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