这段代码无论我多么努力,我都无法理解......
#include <iostream>
using namespace std;
int main()
{
int ***mat;
mat = new int**[4];
for(int h = 0; h < 4; h++) {
mat[h] = new int*[4];
}
for (int i = 0; i < 4; i++) {
delete[] mat[i];
delete[] mat;
}
return 0;
}
不应该mat = new int**[4];
表示mat指向int**
数组,所以当我想使用此数组的成员时,我应该*mat[0]
?
我没有得到这一行mat[h] = new int*[4];
。
答案 0 :(得分:5)
以下是关于内容评论和更正代码的逐步解释。
#include <iostream>
using namespace std;
int main()
{
int ***mat; // placeholder for a 3D matrix, three dimensional storage of integers
// say for 3d matrix, you have height(z-dimension), row and columns
mat = new int**[4]; // allocate for one dimension, say height
for(int h = 0; h < 4; h++) {
mat[h] = new int*[4]; // allocate 2nd dimension, say for rows per height
}
// now you should allocate space for columns (3rd dimension)
for(int h = 0; h < 4; h++) {
for (int r = 0; r < 4; r++) {
mat[h][r] = new int[4]; // allocate 3rd dimension, say for cols per row
}}
// now you have the matrix ready as 4 x 4 x 4
// for deallocation, delete column first, then row, then height
// rule is deallocate in reverse order of allocation
for(int h = 0; h < 4; h++) {
for (int r = 0; r < 4; r++) {
delete [] mat[h][r]; // deallocate 3rd dimension, say for cols per row
}
delete [] mat[h]; // deallocate 2nd dimension, rows per height
}
delete [] mat; // deallocate height, i.e. entire matrix
return 0;
}
答案 1 :(得分:0)
一个级别的指针&#39;隐含的是
new int;
将为您提供int*
,
new int[3];
从那里向上移动