c – 双阵列特里问题

int arr2D[2][3];
int arr1D[2 * 3]; // # of rows * # of columns

// access element [1][2] of 2D array,i.e.,the last element
int elem2D = arr2D[1][2];
// access element [1][2] of 1D array,the last element
int elem1D = arr1D[1 * 3 + 2];
// =========================================================
lets visualize the array access:
arr2D => x/y 0  1  2
          0 [n][n][n]
+1 on x > 1 [n][n][n]
+2 on y ---------- ^
y_len  =>   ^-------^ 3 elements
so the access happens with x * y_len + y
                           1 *   3   + 2
Now for the 1D array
we want the second row,so we go with 1 * 3
(0-based access,y_len = 3)
and then we want the 3rd element,so + 2
(again,0-based access)
arr1D =>  x  0  1  2 
            [n][n][n]
             3  4  5
1 * 3 = 3 > [n][n][n]
      + 2 = 5 ---- ^

我希望我没有让这太复杂(尽管我认为我做了……).