如何通过函数参数返回动态数组？

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;
}

Answer 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)

Answer 2

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 );   

Answer 3

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++)