二维数组和动态分配如何工作？

Question

So my idea:

A 2-D array is an array of arrays. If an array is X by Y, then each X points to an array with length Y. Is that part correct? Is that why also when you want to dynamically allocate an array in C you have to say

array=(type **)malloc(X*sizeof(type *)). 

The type * is to tell the compiler that each array index will be a pointer to another array so it should be sufficiently big enough to hold pointers o array?

Also why is this loop needed?

for(i =0;i<X;i++)
{
array[i]=(type *)malloc(Y*sizeof(type))
}

Is the cast

(type *)

needed because so that each index is a pointer to a 1-D array? But this time in the malloc we can just say (type) instead of (type *) because the index won't be holding pointers any more?

EDIT

Oh..so this is just mimicking a 2-d array by having an each index in an 1-d array point to another 1-d array. Ok. In that case is my logic for why each malloc have that specific argument make sense? The code here is taken from this SO question dynamic allocation/deallocation of 2D & 3D arrays

Answer 1

Your code doesn't involve an actual 2D array. This is a 2D array:

int myArray[X][Y];

It involves no pointers.

Your code involves a pointer to an array of pointers, which isn't the same thing.

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Why is this loop needed?

Because otherwise all of the pointers in your array of pointers don't point at any storage.

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Is the cast needed?

No.