我如何在一个分配C中动态分配2D数组 [英] How can I dynamically allocate 2D-array in one allocate C

问题描述

您能帮我弄清楚如何在一个分配调用中分配2D数组吗?

Can you help me figure out how to allocate a 2D-array in one allocate call?

我试图做:

int ** arr =(int **)malloc(num * num * sizeof(int *));

int** arr = (int**)malloc(num * num * sizeof(int*));

但是它不起作用.

num是行和列.

推荐答案

如何在1个分配C中动态分配array2D

How can i to allocate dynamically array2D in 1 allocate C

让我们从2D数组开始:
2D数组或数组4的数组3 int"

Let us start with what is a 2D array:
Example of a 2D array or "array 3 of array 4 of int"

int arr1[3][4];
arr1[0][0] = this;

OP的代码声明了指向int的指针，而不是2D数组或2D数组的指针.
顺便说一句，不需要强制转换.

OP's code declares a pointer to pointer to int, not a 2D array nor a pointer to a 2D array.
BTW, the cast in not needed.

int** arr = (int**)malloc(num * num * sizeof(int*));

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代码可以为2D数组分配内存，并返回指向该内存的指针. 指向int数组6的数组5的指针

 int (*arr2)[5][6] = malloc(sizeof *arr2);
 if (arr2 == NULL) return EXIT_FAILURE;
 (*arr2)[0][0] = this;
 return EXIT_SUCCESS;

 // or with Variable Length Arrays in C99 and optionally in C11
 int (*arr3)[num][num] = malloc(sizeof *arr3);
 (*arr3)[0][0] = that;

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或者，代码可以为一维数组分配内存，并返回指向该内存的指针. 指向int数组8的指针.有时，这通常是人们想要的带有分配2D"数组的东西，实际上是指向1D数组的指针

---

Alternatively code can allocate memory for a 1D array and return a pointer to that memory. pointer to array 8 of int. Sometimes this is often what one wants with with an "allocate 2D" array, really a pointer to a 1D array

 int (*arr4)[8] = malloc(sizeof *arr4 * 7);
 arr4[0][0] = this;

 // or
 int (*arr5)[num] = malloc(sizeof *arr5 * num);
 arr5[0][0] = that;

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