std::begin and std::end not working with pointers and reference why?

Question

Why std::begin() and std::end() works with array but not pointer[which is almost array] and reference of array [which is alias of original array].

After scratching my head for 15 min i am not able to get anything in google.

Below only first case works, not second and third, what could be the reason for this?

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>

int main() 
{
   int first[] = { 5, 10, 15 };  // Fist Case

    if (std::find(std::begin(first), std::end(first), 5) != std::end(first)) {
        std::cout << "found a 5 in array a!\n";
    }

   int *second = new int[3];  // Second Case
    second[0] = 5;
    second[1] = 10;
    second[2] = 15;
    if (std::find(std::begin(second), std::end(second), 5) != std::end(second)) {
        std::cout << "found a 5 in array a!\n";
    }

    int *const&refOfFirst = first;  // Third Case

        if (std::find(std::begin(refOfFirst), std::end(refOfFirst), 5) != std::end(refOfFirst)) {
        std::cout << "found a 5 in array a!\n";
    }
}

Error:

error: no matching function for call to ‘begin(int&)’
  if (std::find(std::begin(*second), std::end(*second), 5) != std::end(*second)) {
                                  ^

@JamesKanze a pointer is half an array. In the same way that an iterators is half a range :) — alfC
– alfC, Commented Jan 14, 2020 at 18:38

Mike Seymour · Accepted Answer · 2014-11-13 13:09:26Z

13

Given just a pointer to the start of an array, there's no way to determine the size of the array; so begin and end can't work on pointers to dynamic arrays.

Use std::vector if you want a dynamic array that knows its size. As a bonus, that will also fix your memory leak.

The third case fails because, again, you're using (a reference to) a pointer. You can use a reference to the array itself:

int (&refOfFirst)[3] = first;

or, to avoid having to specify the array size:

auto & refOfFirst = first;

and begin and end will work on this exactly as they would work on first itself.

edited Nov 13, 2014 at 13:09

answered Nov 13, 2014 at 13:03

Mike Seymour

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3 Comments

alfC Over a year ago

"Given just a pointer to the start of an array, there's no way to determine the size of the array; so begin and end can't work on pointers to dynamic arrays.", well, there is at least a begin. I have seen generic programming cases in which T* begin(T* ptr){return ptr;} would make sense to exist.

Caleth Over a year ago

@alfC begin without end is asking for trouble.

alfC Over a year ago

@Caleth well, using a (single) pointer to describe a finite range is asking for trouble in the first place (OP question). In fact I don’t define an ‘end’ so you will get a compilation error eventually. My point is that you can still have generic code if what you need is a sentinel (see Rangesv3) rather than a normal end. (Some algorithms applied to an unended range are (can be) guaranteed to terminate even without and ‘end’.)

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