Array Basics (sizeof)

[pYro_65]

But in that situation, the size is known at compile time - it is know to be: sizeof(int)*n - of course it is known, otherwise it wouldn't be able to do malloc() for the array. Clearly during execution it has to do that multiplication.

n isn't known at compile time, only the sizeof int. Also its stored on the stack.

pYro_65:

typedef int arr_t[ 10 ];

And that one is definitely known at compile time as sizeof(arr_t) is a type. Doing that is no different from passing an array to a function defined as:
void someFunc (int array[10]);

Yup, thats what I was showing.

In fact if you were to do this the size ceases to be known:

void someOtherFunc(int* arr){
  sizeof(arr); //=sizeof(pointer)
}
arr_t array = {1,2,3,4,5,6,7,8,9,0};
someOtherFunc((int*)array_t);

Yup, thats exactly what I've pointed out. int*arr is not the array and never will be, it only holds the address of the first element.
I showed clearly above how to pass a pointer to an array and get the correct size using sizeof.

void Foo( arr_t *arrp ){
  
  Serial.println( sizeof( *arrp ) / sizeof( **arrp ) );
}