C++ returning array and display it

Question

i would like to ask, why this code doesnt work...

int* funkce(){
    int array[] = {1,2,3,4,5};
    return(array);
  }

int main(){

    int* pp = funkce();
    int* ppk = pp+5;

    for (int *i = pp; i!=ppk; i++){
        cout << (*i) << endl;
    }

    system("PAUSE");
    return(0);
}

This code display:

So the poniter is out of array... But how is possible, that this code with array in Main is ok?

int main(){

    int a[] = {1,2,3,4,5};
    int* pp = a;
    int* ppk = pp+5;

    for (int *i = pp; i!=ppk; i++){
        cout << (*i) << endl;
    }
    system("PAUSE");
    return(0);
}

this code displayed:

Could you explain to me, why the first one doesnt work? Thank you!

stackoverflow.com/questions/6441218/…

Qaz
– Qaz

2014-10-24 00:04:31 +00:00
Commented Oct 24, 2014 at 0:04 — Qaz
– Qaz, Commented Oct 24, 2014 at 0:04

Barry · Accepted Answer · 2014-10-23 23:52:40Z

6

You're returning a pointer to a temporary which goes out of scope when the function ends. If you want to have a function return an array, you need to do one of:

std::array<int, 5> func() {
    // stack-allocated
    std::array<int, 5> a = {1, 2, 3, 4, 5};
    return a;
}

std::vector<int> func() {
    // heap-allocated
    std::vector<int> a = {1, 2, 3, 4, 5};
    return a;
}

int* func() {
    // heap-allocated, you have to remember to delete it
    int* a = new int[5]{1, 2, 3, 4, 5};
    return a;
}

etc. There's more options, but this should give you a good start.

answered Oct 23, 2014 at 23:52

Barry

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8 Comments

pm100 Over a year ago

std::vector is the one to use (IMHO)

Neil Kirk Over a year ago

@pm100 Why not std::array?

pm100 Over a year ago

because the std:array will return a copy of the array (more expensive in general although probably trivial here). std::vector returns a pointer to the array on the heap. And the perf of vector and array are the same

Barry Over a year ago

@pm100 If you have std::array<int, 5> a = func(), the array won't be copied, it will be constructed in place. The determination to use one or the other is entirely use-case based.

pm100 Over a year ago

according to your own comment it was allocated on the stack of the func. Cant be on the func stack when I return it

|

Deduplicator · Accepted Answer · 2014-10-23 23:51:54Z

0

Never return a local a pointer/reference to a variable, that memory is freed/re-used on return.

Using that dangling reference is Undefined Behavior and has unpredictable consequences.

Return an object, a pointer to heap-allocated memory, or save into a caller-supplied buffer.

answered Oct 23, 2014 at 23:51

Deduplicator

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