以下是在堆上创立2D数组(或动态分配2D数组)的不同办法。
在以下示例中, 咱们思考了”[R‘作为行数, ‘C‘作为列数, 咱们创立了一个2D数组, 其中r = 3, c = 4和以下值
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1)应用单个指针:
一种简略的办法是应用简略的指针算法调配大小为r * c的存储块和拜访元素。
#include <stdio.h>
#include <stdlib.h>
int main()
{
int r = 3, c = 4;
int *arr = ( int *) malloc (r * c * sizeof ( int ));
int i, j, count = 0;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
*(arr + i*c + j) = ++count;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf ( "%d " , *(arr + i*c + j));
/* Code for further processing and free the
dynamically allocated memory */
return 0;
}
输入如下:
1 2 3 4 5 6 7 8 9 10 11 12
2)应用指针数组
咱们能够创立大小为r的指针数组。请留神, 从C99开始, C语言容许应用可变大小的数组。创立指针数组后, 咱们能够为每行动静分配内存。
#include <stdio.h>
#include <stdlib.h>
int main()
{
int r = 3, c = 4, i, j, count;
int *arr[r];
for (i=0; i<r; i++)
arr[i] = ( int *) malloc (c * sizeof ( int ));
// Note that arr[i][j] is same as *(*(arr+i)+j)
count = 0;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
arr[i][j] = ++count; // Or *(*(arr+i)+j) = ++count
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf ( "%d " , arr[i][j]);
/* Code for further processing and free the
dynamically allocated memory */
return 0;
}
输入如下:
1 2 3 4 5 6 7 8 9 10 11 12
3)应用指向指针的指针
咱们还能够应用双指针动态创建指针数组。一旦咱们动态分配了数组指针, 就能够像办法2一样为每行动静分配内存。
#include <stdio.h>
#include <stdlib.h>
int main()
{
int r = 3, c = 4, i, j, count;
int **arr = ( int **) malloc (r * sizeof ( int *));
for (i=0; i<r; i++)
arr[i] = ( int *) malloc (c * sizeof ( int ));
// Note that arr[i][j] is same as *(*(arr+i)+j)
count = 0;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
arr[i][j] = ++count; // OR *(*(arr+i)+j) = ++count
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf ( "%d " , arr[i][j]);
/* Code for further processing and free the
dynamically allocated memory */
return 0;
}
输入如下:
1 2 3 4 5 6 7 8 9 10 11 12
4)应用双指针和一个malloc调用
#include<stdio.h>
#include<stdlib.h>
int main()
{
int r=3, c=4, len=0;
int *ptr, **arr;
int count = 0, i, j;
len = sizeof ( int *) * r + sizeof ( int ) * c * r;
arr = ( int **) malloc (len);
// ptr is now pointing to the first element in of 2D array
ptr = ( int *)(arr + r);
// for loop to point rows pointer to appropriate location in 2D array
for (i = 0; i < r; i++)
arr[i] = (ptr + c * i);
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
arr[i][j] = ++count; // OR *(*(arr+i)+j) = ++count
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf ( "%d " , arr[i][j]);
return 0;
}
输入如下:
1 2 3 4 5 6 7 8 9 10 11 12
如果发现任何不正确的中央, 或者想分享无关上述主题的更多信息, 请发表评论。
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