C++ function returning reference to array

Question

Is there any other way to receive a reference to an array from function returning except using a pointer?

Here is my code.

int ia[] = {1, 2, 3};
decl         


        

Answer 1

Your functions signature would look like this:

int(&A::adapter(/*input_args_or_empty*/))[3]{
   return ia;
}

Answer 2

Some alternatives I think are superior in C++14 onwards:

auto and array reference trailing return type:

template <typename T, std::size_t N>
auto array_ref_test(T(&array)[N]) -> T(&)[N]
{
    return array;
}

decltype(auto):

template <typename T, std::size_t N>
decltype(auto) array_ref_test(T(&array)[N])
{
    return array;
}

Example usage:

int array_ref_ [] = { 1, 2, 3, 4, 5 };   
decltype(auto) array_ref_result = array_ref_test(array_ref_); //'decltype(auto)' preserves reference unlike 'auto'
std::cout << std::size(array_ref_result);

Answer 3

Is there any other way to receive a reference to an array from function returning except using a pointer?

Yes, using a reference to an array, like with any other type:

int (&ref)[3] = a.foo();

To avoid the clunky syntax, you could use a typedef instead.

typedef int int_array3[3];

...
int_array3& foo() { return ia; }

...

int_array3& ref = a.foo();

Answer 4

You should use std::array to avoid confusion and make cleaner, safer and less clunky code:

class A {
public:
    typedef std::array<int,3> array;
    A() : ia{1, 2, 3} {}
    array &foo(){ return ia; }
private:
    array ia;
};

int main() {
    A a;
    auto i1 = a.foo();      // ok, type of i1 is A::array, it is a copy and visit array by i1[0]
    for ( int i : i1 ) {}   // you can iterate now, with C array you cannot anymore
    auto &i2 = a.foo();     // ok, type of i2 is A::array&, you can change original by i2[0] = 123 
    A::array i3 = a.foo();  // fine, i3 is now a copy and visit array by i3[0]
    A::array &i4 = a.foo(); // fine, you can change original by i4[0] = 123
    int *i5 = a.foo().data();  // if you want old way to pass to c function for example, it is safer you explicitly show your intention
    return 0;
}

I know the name of the array is a pointer to the first element in that array

This is incorrect, array can be implicitly converted to a pointer to the first element. It is not the same.

Answer 5

To answer this question one can design a solution around std::reference_wrapper. This code compiles and runs.

#pragma once
#include <array>
#include <functional>
#include <cstdlib>
#include <ctime>

namespace {

/// <summary>
/// some generic utility function 
/// with argument declared as 
/// native array reference
/// for testing purposes only
/// </summary>
template<typename T, std::size_t N >
inline void array_util_function( T (&arr_ref) [N] ) {
    auto random = [](int max_val, int min_val = 1) -> int {
        // use current time as seed for 
        // random generator, but only once
        static auto initor = []() { 
            std::srand((unsigned)std::time(nullptr)); return 0; 
        }();

        return min_val + std::rand() / ((RAND_MAX + 1u) / max_val);
    };

    for (auto & element : arr_ref) {
        element = random(N);
    }
}

// the key abstraction
template <typename T, std::size_t N>
using native_arr_ref_wrapper = std::reference_wrapper<T[N]>;

template<typename T, std::size_t N >
constexpr std::size_t array_size(T(&arr_ref)[N]) {
    return N;
}

// return the size of the native array
// contained within
template<typename T, std::size_t N >
constexpr std::size_t array_size(native_arr_ref_wrapper<T,N> & narw_) {
    return array_size(narw_.get());
}

/// <summary>
/// returns std::reference_wrapper copy
/// that contains reference to native array 
/// in turn contained inside the std::array
/// argument
/// </summary>
template<typename T, std::size_t N >
    inline auto
        native_arr_ref(const std::array<T, N> & std_arr)
        -> native_arr_ref_wrapper<T,N>
{
    using nativarref = T(&)[N];

    return std::ref(
            (nativarref)*(std_arr.data())
        );
}

    /// <summary>
    /// returns std::reference_wrapper copy
    /// that contains reference to native array 
    /// </summary>
    template<typename T, std::size_t N >
    inline auto
        native_arr_ref(const T (& native_arr)[N] )
        -> native_arr_ref_wrapper<T, N>
    {
        using nativarref = T(&)[N];

        return std::ref(
            (nativarref)*(native_arr)
        );
    }

    auto printarr = [](auto arr_ref, const char * prompt = "Array :") {
        printf("\n%s:\n{ ", prompt);
        int j = 0;
        // use the array reference 
        // to see the contents 
        for (const auto & element : arr_ref.get()) {
            printf(" %d:%d ", j++, element);
        }
        printf(" }\n");
    };

// arf == native array reference
void test_ () {

using std::array;
// the arf type is 
// std::reference_wrapper<int[10]> 
auto arf  = native_arr_ref(array<int, 10>{});
// the arf type is same here 
// but made from a native array
auto ar2 = native_arr_ref({ 0,1,2,3,4,5,6,7,8,9 });
printarr(ar2);

// notice how argument to native_arr_ref()
// does not exist here any more
// but, wrapper contains the copy 
// of the native array  

// type is here: int[10]&
auto & native_arr_ref = arf.get();

// due to array type decay 
// the type is here: int *
auto native_arr_ptr = arf.get();

// size_t
const auto size = array_size(arf);

// this function requires
// native array as argument
// this is how we use arf in place of 
// native arrays
array_util_function(arf.get());

printarr(arf);

    }
  }