pointers frustration

[4ET0pfjN]

Hi, I don't want pointers to frustrate me, but I was sure this was correct, which just empties an array for a given array and size passed as parameters.

int *emptyList(int *list, int arraySize)
{
	for ( int i = 0; i < arraySize; i++)
	{
		emptyList[i] = 0;
	}
		
	return *emptyList;//NB: WHY isn't it: return *emptyList
}

It works when I replace with:
return emptyList;

I know that a pointer to an array is always pointer to the first elem only, a constant pointer so it should work...I don't want pointers to haunt me if I am going to be a developer as a career.

[Peter87]

emptyList is the name of the function, but you try to use it as an array.

[4ET0pfjN]

that was a typo in my code, I meant

int *emptyList(int *list, int arraySize)
{
	for ( int i = 0; i < arraySize; i++)
	{
		emptyList[i] = 0;
	}
		
	return *list;//NB: WHY can't it be: return *list
}

Why is return *list not valid, it's just a pointer to the FIRST elem of this array?

[1yR4jE8b]

Your function return value is int* ( a pointer to an int -- whether it's an array or a single one).

list is a pointer
*list is the value pointed at by the pointer.

[4ET0pfjN]

So we use return list because our pointer is pointing to set of memory addresses b/c the compiler knows we will be passing an array in the parameter, and if I use
return *list I am referring to just the first element value, is that right?

[Pravesh Koirala]

Your initial code

int *emptyList(int *list, int arraySize)
{
	for ( int i = 0; i < arraySize; i++)
	{
		emptyList[i] = 0;
	}
		
	return *emptyList;//NB: WHY isn't it: return *emptyList
}

Corrected code

int *emptyList(int *list, int arraySize)
{
	for ( int i = 0; i < arraySize; i++)
	{
	         list[i] = 0;         //Modified here. emptyList is a function. list is an integer pointer
	}
		
	return list;   // so, after you have emptied it, you do this to return it. because list is an integer pointer. emptyList is a function
}