Assign value to a pointer array give me error while using pointer arithmetic

Question

When I do this pointer arithmetic , it always give me an error


int main()
{
    const int rowSize=40;
    int* unique=nullptr;

    int arr[rowSize]={1,1,11,31,21,22,2,2,3,32,31,3,4,34,45,5,55,5,55,5,6,46,64,6,7,27,74,7,7,7,7,11,11,11,11,11,1,2,13,4};
    int amount=0;

    for (int count=0; count<rowSize; count++)
    {
        if (arr[count]!=arr[count+1])
        {
            amount++;
        }
    }
    unique= new int[amount];
    for (int count=0; count<rowSize-1; count++)
    {
        if (arr[count]!=arr[count+1])
        {
            *unique=arr[count];
            unique++;
        }
    }

    for (int count=0; count<20; count++)
    {
        cout<<unique[count]<<" ";
    }
    cout<<endl;
    delete [] unique;
    unique=nullptr;
    return 0;
}

Everytime I do this pointer arithmetic, *unique=arr[count] and unique++, it always give me a funky output at the end.

int the line if( arr[count] != arr[count+1]) when count is on the last element (rowSize -1) you access arr[count+1] where count+1 == rowSize this is out of range of your array. The last element is then just some garbage value ;) just make your loop for (int count=0; count < rowSize - 1; count++) and everthing will work. — Dimfred
– Dimfred, Commented May 5, 2020 at 13:59
@cigien MB, rowSize=40 I forgot to put it during copying this code. it is just a piece of function, it compile thou but it just give a funny output. — prayoga yoga
– prayoga yoga, Commented May 5, 2020 at 14:08
You might think that the information you leave out is not relevant. But we cant know that, so you should always provide a minimal-reproducible-example — cigien
– cigien, Commented May 5, 2020 at 14:10

Adrian Mole · Accepted Answer · 2020-05-05 14:41:52Z

1

Changing the value of a pointer returned by the new[] operator is extremely dangerous, and never really the right way to go about coding. This is because, at some point, you will need to release that memory with a delete[] 'call' to the address that new[] gave you.

In your case, the module that calls your uniqueArr function gets a returned value that will no longer (in most cases) be the correct address for the delete[] call, and that will fail.

You would be far better off using the [] operator on your pointer, using an index value that you increment when appropriate (where you currently increment the pointer). Something like this:

//...
    int* unique = new int[amount];
    size_t uIndex = 0;

    for (int count=0; count<rowSize-1; count++)
    {
        if (arr[count]!=arr[count+1])
        {
            unique[uIndex] = arr[count]; // The [] works fine with a 'new[]' pointer!
            uIndex++; // You could even include this POST-increment inside the [], above!
        }
    }
//... You can now 're-use' "unique" as it will still be the original address.
//...

In this way, the value returned by the function will be unchanged from that returned by the new[] operator, and will be valid for any subsequent delete[] operation.

Feel free to ask for further clarification an/or explanation.

EDIT: An alternative (though, IMHO not good) approach would be to have a second int* pointer in which you save the value of unique, then restore unique to that value before 're-using' it (in a second loop, or before calling delete[]).

edited May 5, 2020 at 14:41

answered May 5, 2020 at 14:31

Adrian Mole

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1 Comment

prayoga yoga Over a year ago

Well, thanks for the help it give me the right output.

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