Whats the difference? Pointer to an array vs regular array

Question

I\'m familiar with Java and trying to teach myself C/C++. I\'m stealing some curriculum from a class that is hosting their materials here. I unfortunately can\'t ask the tea

Answer 1

My understanding is that an array in C is simply a reference to the memory address of the first element in the array.

So, what is the difference between int *pointerArray = new int[10]; and int array[10]; if any?

What you mention is the reason for much confusion in any C/C++ beginner.

In C/C++, an array corresponds to a block of memory sufficiently large to hold all of its elements. This is associated to the [] syntax, like in your example:

int array[10];

One feature of C/C++ is that you can refer to an array by using a pointer to its type. For this reason, you are allowed to write:

int* array_pointer = array;

which is the same as:

int* array_pointer = &array[0];

and this allows to access array elements in the usual way: array_pointer[3], but you cannot treat array as a pointer, like doing pointer arithmetics on it (i.e., array++ miserably fails).

That said, it is also true that you can manage arrays without using the [] syntax at all and just allocate arrays by using malloc and then using them with raw pointers. This makes the "beauty" of C/C++.

Resuming: a distinction must be made between the pointer and the memory that it points to (the actual array):

1. the [] syntax in declarations (i.e., int array[10];) refers to both aspects at once (it gives you, as to say, a pointer and an array);

2. when declaring a pointer variable (i.e., int* p;), you just get the pointer;

3. when evaluating an expression (i.e., int i = p[4];, or array[4];), the [] just means dereferencing a pointer.

Apart from this, the only difference between int *pointerArray = new int[10]; and int array[10]; is that former is allocated dynamically, the latter on the stack.