Using iterators on arrays

An array is converted to a pointer easily, but not always. For example, if you take the address of the array or get a reference then the original array type isn't lost:

int a[10];int (&ar)[10] = a; // fineint (*ap)[10] = &a; // also fine

However, when you use an array in a way where most other types would be copied, the array is converted to a pointer and the pointer is copied instead.

In your example, you can use arr2 if you make it be a reference:

 auto &arr2 = arr;

Now arr2 has type int (&)[10] instead of int *.

Because std::begin is defined for arrays but not for pointers. An array type is not the same as a pointer type.

The problem here is that apparently the auto arr2 = arr degrades the type from the array type to a pointer type.

Note that the problem is actually in std::end not in std::begin. After all how would std::end be able to give the pointer to the last+1 element when all it has is a pointer to the first element? It can't, therefore std::end cannot be defined for a pointer type and hence std::begin does not make any sense for a pointer type.

To be precise the type of arr is int[10] while that of arr2 and arr3 is int*. The former can degrade into the latter but not vice versa.

Here, auto will decay the arr (array type) to a pointer. And both std::begin and std::end can only work for containers or arrays, but not for pointers.

From C++11 standard

§7.1.6.4 auto specifier [dcl.spec.auto]

1 The auto type-specifier signifies that the type of a variable being declared shall be deduced from its initializeror that a function declarator shall include a trailing-return-type.

You code cannot work here as auto cannot deduce array types. Just like:

char a[5];auto b[5] = a;  // error, a evaluates to a pointer, which does                // not match the array type

To make it work, just use C++ containers like std::vector:

vector<int> arr = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };