如何使用指针访问N-d数组元素

Supposing we have a 2-d array a[n][m]. Instead of a[i][j] I can use *(*a + i * n + j).

假设我们有一个二维数组a [n] [m]。而不是[i] [j]我可以使用*(* a + i * n + j)。

How can I access an element of N-dimensional array using pointers? i.e What can I use instead of a[i][j]...[k]?

如何使用指针访问N维数组的元素?即我可以使用什么而不是[i] [j] ...... [k]?

For example, I tried doing the following, but this doesn't work:

例如,我尝试执行以下操作,但这不起作用:

#include<iostream>
#include<conio.h> 
using namespace std;
int main() {
    const int n1 = 5, n2 = 5, n3 = 5;
    int array[n1][n2][n3];
    for (int i = 0; i < n1; i++)
    for (int j = 0; j < n2; j++)
    for (int k = 0; k < n3; k++)
        array[i][j][k] = n1*n2*n3;

    cout << array[1][2][4]<<endl;
    cout << *(*array + n1 + n2 * 2 + 4);

    _getch();
    return 0;
}

1 个解决方案

#1

Quote from N1256 6.5.2.1 Array subscripting

引自N1256 6.5.2.1数组下标

The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))).

下标运算符[]的定义是E1 [E2]与(*((E1)+(E2)))相同。

From this, I can say that all of the followings are equivalent:

由此可以说,以下所有内容都是等效的:

a[i][j][k]
(*(a + i))[j][k]
(*((*(a + i)) + j))[k]
(*((*((*(a + i)) + j)) + k))

Note: The added code is C++, not C. For C++, N3337 5.2.1 Subscripting says that:

注意:添加的代码是C ++,而不是C.对于C ++,N3337 5.2.1 Subscripting说:

The expression E1[E2] is identical (by definition) to *((E1)+(E2))

表达式E1 [E2]与*((E1)+(E2))相同(根据定义)

#1