嗯,这是一个完整的样本,但是在最后一次打印后控制台消失了,我不能让它留下来。我还在一些行中包含了一些查询
//bidimensional array dynamic memory allocation
#include <stdio.h>
#include <stdlib.h>
void main()
{
int **p; // pointer to pointer
int n,m,i,j,k; // n is rows, m is cols, i and j are the indexes of the array, k is going to be like m, but used to print out
do
{
printf("\n how many rows?");
scanf ("\%d", &n);
}
while (n <= 0);
//为n个元素的数组预留内存,每个元素都是一个指向int(int *)
的指针//查询:指向int的指针?不是指向指针的指针吗?它使用**
p = (int **) malloc (n * sizeof(int *)); //
if(p == NULL)
{
printf("Insuficient memory space");
exit( -1);
}
for (i = 0; i < n; i++) // now lets tell each row how many cols it is going to have
{
printf("\n\nNumber of cols of the row%d :", i+1); // for each row it can be different
scanf("%d", &m); // tell how many cols
p[i] = (int*)malloc(m * sizeof(int)); // we allocate a number of bytes equal to datatype times the number of cols per row
/ 查询:我无法掌握p [i],因为如果p是指针的指针,那个数组符号是什么,我的意思是方括号 /
if(p[i] == NULL)
{ printf("Insuficient memory space");
exit(-1);
}
for (j=0;j<m;j++)
{
printf("Element[%d][%d]:", i+1,j+1);
scanf("%d",&p[i][j]); // reading through array notation
}
printf("\n elements of row %d:\n", i+1);
for (k = 0; k < m; k++)
// printing out array elements through pointer notation
printf("%d ", *(*(p+i)+k));
}
// freeing up memory assigned for each row
for (i = 0; i < n; i++)
free(p[i]);
free(p);// freeing up memory for the pointers matrix
getchar(); // it cannot stop the console from vanishing
fflush(stdin); // neither does this
}
// * ** * **** 非常感谢 * ** * **
答案 0 :(得分:2)
很容易理解数组上下文中的指针。 因此,如果
int * p是int的一维数组,那么int ** p将是int的二维数组。换句话说,它是一个包含指向一维数组的指针的数组。
所以
p = (int **) malloc (n * sizeof(int *)); //是指向
的指针
和p [i]是指向int的当前指针。