Is it possible to use iterator in C array? - c++

Question

I know iterator can be used for vector either with std::vector::begin or with std::begin defined in <iterator>. Same for std::end.
Can I also use iterators with C arrays? I tried the following but it didn't work.
#include <iostream>
#include <iterator>
using std::cin;
using std::cout;
using std::endl;
using std::begin;
using std::end;
void print(const int *arr) {
for (auto cbeg = cbegin(arr); cbeg != cend(arr); ++cbeg) {
cout << *cbeg << endl;
}
}
int main() {
int arr[] = {9, 18, 31, 40, 42};
print(arr);
}
edit:
I thought i could do this because of this piece of code in C++ primer where they used begin and end to get iterators to first and to one past the end element:
#include <iterator>
using std::begin; using std::end;
#include <cstddef>
using std::size_t;
#include <iostream>
using std::cout; using std::endl;
// const int ia[] is equivalent to const int* ia
// size is passed explicitly and used to control access to elements of ia
void print(const int ia[], size_t size)
{
for (size_t i = 0; i != size; ++i) {
cout << ia[i] << endl;
}
}
int main()
{
int j[] = { 0, 1 }; // int array of size 2
print(j, end(j) - begin(j));
return 0;
}

Answer 1

Yes, iterators can be used on arrays. For example
int main()
{
int arr[] = {9, 18, 31, 40, 42};
for (auto cbeg = cbegin(arr); cbeg != cend(arr); ++cbeg) {
cout << *cbeg << endl;
}
will print all elements of arr.
The problem is that, your print() function accepts a pointer, so iterators cannot be used in that case.
However, if you change print() to
void print(int(&arr)[5])
{
// same body as before
}
or (if you don't want the size fixed as 5) to
template<int N> void print(int(&arr)[N])
{
// same body as before
}
you will find it will work, since the array is passed by reference. Note that these functions will not compile if pointers are passed to them.

Answer 2

You can use std::begin and std::end on c-style arrays. You can't use them on pointers. C-style arrays decay to pointers when passed to functions unless you pass them by reference. Dynamic arrays (allocated by new) are also accessed via a pointer, so that doesn't work either.
Passing by reference like void print(int (&arr)[5]) {...} should work. Remark that you need templates if you want variable sized arrays.

Answer 3

You seem to be looking for
// use array pointer to dodge array decay of function parameters:
void obfuscated_meta_programming (int (*arr)[5])
{
for(auto it = std::begin(*arr); it != std::end(*arr); it++)
{
cout << *it << endl;
}
}
int main()
{
int arr[] = {9, 18, 31, 40, 42};
obfuscated_meta_programming(&arr);
}
Which can be rewritten in a sane way like this:
void print (int arr[5])
{
for(size_t i=0; i<5; i++)
{
cout << arr[i] << endl;
}
}
int main()
{
int arr[] = {9, 18, 31, 40, 42};
print(arr);
}

Answer 4

Not this way. Your function print gets only const int * pointer. This pointer has no information about the length of array arr[] from function main, so if you have pointer to int as parameter, then even theoretically, there is no way you could iterate in function "print", regardless whether you use iterator or not. One way to go would be to use std::array and initialize it with char[], like this:
#include <iterator>
#include <iostream>
#include <array>
using std::cin;
using std::cout;
using std::endl;
template<size_t N> void print(std::array<int,N> &arr) {
for (const auto &s : arr) {
cout << s << endl;
}
}
int main() {
std::array<int,5> arr = {9, 18, 31, 40, 42};
print(arr);
}
Compile with --std=c++11

Answer 5

C arrays are not length prefixed. If you have the size, you can create a pointer which satisfies Iterator requirements:
begin = arr;
end = arr + size;
some_algorithm(begin, end);