How does this compile

[EwCjE3v7]

The following code has a problem, but compiles and runs fine

int &get(int *arry, int index)
{
	return arry[index];
}

int main()
{
	int ia[10];
	for (int i = 0; i != 10; ++i) {
		get(ia, i) = i;
		cout << ia[i] << " ";
	}
	return 0;
}

PROB : We are passing ia, to *arry which would yield a pointer and not the array.

Maybe I`m wrong, what I learnt was that when we pass an array we are actually passing a pointer to the first element. Maybe I`m wrong so please tell me if I am

[keskiverto]

I learnt was that when we pass an array we are actually passing a pointer to the first element.

Yes. Therefore, you pass a pointer to the function.

The function stores that pointer in a pointer variable 'arry'. Then it uses that pointer in a normal way.

When you have a pointer, you can dereference it. One syntax is *arry, but that is equivalent to *(arry+0) and in turn equivalent to arry[0]. All three refer to value in the address that the pointer has.

'arry' is an 'int' pointer. If you take the address and add to it, say *(arry+7), you don't actually add '7' to the address, but the number of bytes that it takes to hold 7 int values.

int a[7];
int * b = &(a[2]);
// a[5] and b[3] refer to same element 

Did that help?

[AbstractionAnon]

We are passing ia, to *arry which would yield a pointer and not the array.

The pointer to the first element and the array are the same thing.
Consider the following statements:

 
  int * iap1 = ia;  // pointer to base of array
  int * iap2 = &ia[0];  // pointer to first element 

iap1 and iap2 are exactly the same.

[EwCjE3v7]

Thank you both, yes that helped a lot. All this time I was confused about when a pointer points to the first element. Thought you could only access that element. Thank you