In C++ we can write:
int arr[] = {20,3,2,0,-10,-7,7,0,1,22};
//Smal Note: Why int *arr = {20,3,2,0,-10,-7,7,0,1,22}; won't work? I learnt I can replace [] with *
but what if I want to allocated arr on heap in one line?
I tried:
int arr[] = new int {20,3,2,0,-10,-7,7,0,1,22};
In function arguments, int[] means "an array of unknown size", and is technically equivalent to int*. (when the pointer is interpreted as a pointer to the first integer in an array).
But in your declaration, int[] means "an array whose size is determined by the initializer", and that is a very well-known size.
new int[] does create an array on the heap, but it returns a pointer to the first element. You might notice a similarity here with function arguments - it's easy to convert from an array to a pointer.
std::vector<int> creates an array on the heap too, but the vector object which manages the array can live anywhere. That's often a lot more convenient.
If you write int arr[] = {20,3,2,0,-10,-7,7,0,1,22}; your arr is usually stored on the stack, just like int a=20, b=3, ...;. In this case, the right-hand side is an initializer, which just tells how int arr[] is initialized.
On the other hand, if you write int *arr = new int[]{20,3,2,0,-10,-7,7,0,1,22}; the array is created on the heap, which can only be accessed via pointers, and the pointer pointing to the array is assigned into int *arr which in turn is on the stack.
So in the context of declaration, int[] and int* are completely different things. But both int[] array and int* array can be accessed using the [] operator, e.g. arr[2] which is synonymous with *(arr+2). And when you use int[] or int* as an argument of a function, they are completely interchangeable.
int *arr = new int[]{...};.std::vector<int>