How can I return dynamic array by function parameter?

Question

How can I return values from my_func by parameter? Application crashes during printf. I don't know why, I thought that *abc will be pointer to xxx...

void my_func(int *_return)
{
    int *xxx = new int[5];
    for (int i = 0; i < 5; i++) xxx[i] = 100+i;

    _return = xxx;

    return;
}

int _tmain(int argc, _TCHAR* argv[])
{
    int *abc = NULL;

    my_func(abc);
    printf("%d", abc[2]);

    return 0;
}

Change every occurrence of _return to *_return (yes, even where you already have one * preceding it). In addition, change my_func(abc) to my_func(&abc). — barak manos
– barak manos, Commented Dec 14, 2014 at 16:10

Vlad from Moscow · Accepted Answer · 2014-12-14 16:08:48Z

1

There are two ways. The first one is to use references. For example

void my_func(int * &a)
{
    a = new int[5];
    for (int i = 0; i < 5; i++) a[i] = 100+i;
}

and the function is called like

my_func( abc );

The second one is to use pointer to pointer. For example

void my_func(int **a)
{
    *a = new int[5];
    for (int i = 0; i < 5; i++) ( *a )[i] = 100+i;
}

and the function is called like

my_func( &abc );

In the both cases you should call

delete [] abc;

when the array will not be needed any more.

Of course you could use std::vector instead of the array

void my_func( std::vector<int> &v )
{
    v.reserve( 5 );
    for (int i = 0; i < 5; i++) v.push_back(  100+i );
}

and the function could be called like

std::vector<int> abc;

my_func( abc );

edited Dec 14, 2014 at 16:08

answered Dec 14, 2014 at 16:00

Vlad from Moscow

313k27 gold badges204 silver badges358 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

mateusz_s Over a year ago

Yes, I was looking for that solution, thank You! It's working, I used the second solution...

mateusz_s Over a year ago

You gave me very detailed solutions, thank You again... especially for the pointer-to-pointer... have a nice day! :)

ravi · Accepted Answer · 2014-12-14 16:03:12Z

1

for (int i = 0; i <= 5; i++) xxx[i] = 100+i;  <<<< i < 5

index is till 4, as 5 would be out of bound.

To make it effective you should pass the address of pointer:-

my_func(&abc);

void my_func(int**_return)

edited Dec 14, 2014 at 16:03

answered Dec 14, 2014 at 15:58

ravi

10.7k1 gold badge21 silver badges36 bronze badges

1 Comment

barak manos Over a year ago

@mateusz_s: Also change _return = xxx to *_return = xxx.

Cory Kramer · Accepted Answer · 2014-12-14 15:58:12Z

0

You have undefined behavior

for (int i = 0; i <= 5; i++)

The only valid indexes for an array of length 5 are [0] to [4], change your termination condition to

for (int i = 0; i < 5; i++)

answered Dec 14, 2014 at 15:58

Cory Kramer

119k19 gold badges176 silver badges233 bronze badges

Collectives™ on Stack Overflow

How can I return dynamic array by function parameter?

3 Answers 3

2 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related