Why my I can't return an array?

Question

I'm trying to return a pointer to an array from a function but I have an issue. When I try to output like this:

#include <iostream>

using namespace std;

int* Somma_Array(int[],int[],int);

int main()
{
    int n;
    cin>>n;
    int A[n],B[n];
    for(int i=0;i<n;i++)cin>>A[i];
    for(int i=0;i<n;i++)cin>>B[i];
    int *c=Somma_Array(A,B,n);
    for(int i=0;i<n*2;i++)cout<<c[i];
}

int* Somma_Array(int v[],int p[],int size)
{
    int r[size*2];
    for(int i=0;i<size;i++)r[i]=v[i];
    for(int i=0;i<size;i++)r[i+size]=p[i];
    return r;
}

it prints weird numbers instead of the actual number. I tried to do what this question says but it does not work. It gives me the following warning:

[Warning] address of local variable 'r' returned [enabled by default]

I'm using bloodshed dev-c++.

http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope — chris, Mar 14 '14 at 19:15
You're returning a pointer to an array that no longer exists as soon as you return. — David Schwartz, Mar 14 '14 at 19:16
Eric Lippert had an [awesome answer](http://stackoverflow.com/a/6445794/392044) for this question.. — Mike Christensen, Mar 14 '14 at 19:16
Also know that `int A[n], B[n]` where `n` is not a compile time constant is a language *extension* supported by some compilers and is not standard. — crashmstr, Mar 14 '14 at 19:17
Because `r` is locally allocated in the call stack, you cannot be passed as parameter to the caller. — Havenard, Mar 14 '14 at 19:17
That r is done (invalid) after return, and a pointer is not holding data (it is just pointing to data) — , Mar 14 '14 at 19:17

score 3 · Answer 1 · answered Mar 14 '14 at 19:17

3

You define a stack allocated array r, which is destroyed when you exit the function Soma_Array. This is one of the (many) reasons vectors are preferred to plain arrays - they handle allocation and deallocation for you.

#include <vector>

std::vector<int> getArray() 
{
    std::vector<int> a = {1, 2, 3};
    return a;
}

answered Mar 14 '14 at 19:17

Engineer2021

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Short form: `return {1, 2, 3};` – chris Mar 14 '14 at 19:22

David Peterson · Answer 2 · 2014-03-14T19:33:55.067

The following:

int r[size*2];

defines r locally. When the function exits (as in the scope of the function expires), r will be destroyed since it is bound to the function's scope. You are likely seeing junk data from the stack frame.

you could fix this by doing the following:

int* r = new int[size * 2];

The variable r will now be heap allocated and exist beyond the scope of the function.

IMPORTANT by doing this, you now must manually free r when you are done with it. So for instance, your calling code will look something like this:

int* result = Somma_Array(v, p, size);
/* ... do stuff ... */
delete[] result;

Since r is an array, note the use of delete[] instead of delete. delete[] is the correct way to destroy arrays.

A Better Alternative

Would std::vector be more what you are after? This is a much safer alternative to hand-rolled arrays. The vector is safer to use, scales automatically as you add elements, and cleans itself up nicely when it leaves scope (assuming you are using a value-type instance). Additionally, vectors can be copied and moved out of functions easily.

score 1 · Answer 3 · answered Mar 14 '14 at 19:16

1

You cannot return arrays in C++. Especially, you should not return a pointer to a local array. You can however return a std::vector<int>:

std::vector<int> Somma_Array(int v[], int p[], int size)
{
    std::vector<int> r(2 * size);
    std::copy(v, v + size, r.begin());
    std::copy(p, p + size, r.begin() + size);
    return r;
}

answered Mar 14 '14 at 19:16

fredoverflow

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How so? The question I linked does it, doesn't it? – olinarr Mar 14 '14 at 19:17
@NetHacker, It returns a pointer (or an `int` depending on which declaration you're talking about). [This](http://coliru.stacked-crooked.com/a/997e6debe4fe3ad8) returns an array. – chris Mar 14 '14 at 19:19
Ok, I'll return a vector so. – olinarr Mar 14 '14 at 19:19
Yes and no. You can, but you can't ;-) You do return an address, not an array ("array decay"). The address points to a deceased local var, as Fred correctly pointed out. That's bad. That's why you can't. You _can_ new an array in that func and return _that_ address, and it will work. – Peter - Reinstate Monica Mar 14 '14 at 19:21

Why my I can't return an array?

3 Answers3