Is there a way of casting a pointer to an array type?

Question

C arrays are somewhat difficult to understand syntactically in C++ and can take some getting used to. Although a 1D array decays to a pointer:

void fn1(int x[2]) {}
void fn2(int*x) {}

fn1() and fn2() have the same function signature.

An array actually does have a type that includes how many elements are in the array. As in:

void fn(int (&)[2]) {}

fn() will only accept a 2 element int array.

Thing is, I can only see that the array of a fixed number of elements can only be generated by stack, file scope or struct/class allocation with that signature:

int twoElementArray[2];

If I were to dynamically allocate it on the heap, I can't seem to get the same signature. I thought that I might be able to cast it, but without success:

int (&array)[2] = reinterpret_cast<int(&)[2]>(new int[2]); // FAIL!

Any ideas as to how this might be accomplished if at all?

EDIT: Although I selected an answer, it actually doesn't actually cast anything, but uses a definitely better method then casting (better not to cast if not required IMO). However, it technically doesn't answer the question since the question asked if there's "a way of casting a pointer to an array type?" The answer is yes.

int (&array)[2] = *reinterpret_cast<int(*)[2]>(new int[2]); // SUCCESS!

Note that I don't necessarily recommend doing this, but it does answer the question. If I needed to convert a pointer to an array type though, that would be how to do it. Read the picked answer for a better solution when using operator new[].

@Infested: Sorry, I don't understand what you mean. Could you explain further and perhaps add an example? — Adrian
– Adrian, Commented Jun 23, 2013 at 8:06
if fn1 only accepts a fixed size of the array, then simply make it a template function and it should get two things , int[] , int FIXED_SIZE = 2, then you can have int x[FIXED_SIZE] — Ariel Pinchover
– Ariel Pinchover, Commented Jun 23, 2013 at 8:10
@Infested: fn1() was part of my explanation as to how arrays work in C++ for the uninitiated. Yes, I could do: void fn3(int* array, size_t FIXED_SIZE) {} or alternatively void fn3(int[] array, size_t FIXED_SIZE) {} which is a common workaround. However, I was trying to utilise the type system to stop the decoupling of the array from it's size. — Adrian
– Adrian, Commented Jun 23, 2013 at 8:24

jrok · Accepted Answer · 2013-06-23 08:34:32Z

10

If I understand your problem correctly, you'd want to do something like this:

// allocate an array of one int[2] dynamically
// and store a pointer to it
int(*p)[2] = new int[1][2];

// now initialize a reference to it
int(&array)[2] = *p;

// delete the array once you no longer need it
delete[] p;

answered Jun 23, 2013 at 8:34

jrok

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Comments

Mark Tolonen · Accepted Answer · 2013-06-23 08:39:17Z

1

I think this is what you are looking for. For the heap, a two-dimensional array int[M][N] decays to int(*)[N]. To pass it by reference, dereference it (see m below):

#include <iostream>
using namespace std;

void func(int (&x)[2])
{
    cout << x[0] << ' ' << x[1] << endl;
}

int main()
{
    // on the heap
    auto m = new int[1][2];
    m[0][0] = 1; m[0][1] = 2;
    auto n = new int[1][3];
    n[0][0] = 4; n[0][1] = 5; n[0][2] = 6;

    // on the stack
    int o[2] = {7,8};
    int p[3] = {9,10};

    func(*m);
    //func(*n); // doesn't compile, wrong size
    func(o);
    //func(p); // doesn't compile, wrong size
}

Output:

1 2
7 8

edited Jun 23, 2013 at 8:39

answered Jun 23, 2013 at 8:33

Mark Tolonen

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1 Comment

Adrian Over a year ago

+1 for using the proper terminology decays. I corrected my question to use the same terminology.

Gian Lorenzo Meocci · Accepted Answer · 2013-06-23 10:02:06Z

-1

Based on Mark answer, this could be better:

template <typename T>
void func(const T &x)
{
    cout << x[0] << ' ' << x[1] << endl;
}

the only bad think is that this code is valid:

cout << x[3] << endl;

answered Jun 23, 2013 at 10:02

Gian Lorenzo Meocci

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1 Comment

Adrian Over a year ago

Sorry, this answer is just wrong. You have just made a template function that takes a const ref of any type. It doesn't take an array at all. Go and try to compile it. You probably meant something along the lines of template <typename T> void func(const T (&x)[2]) {...} or template <typename T, size_t N> typename std::enable_if<(N>=2)>::type func(const T (&x)[N]) {...}. However, that's also assuming that I want to take in any type and not just an int, which isn't what the question asked.

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