pointers

Mar 20, 2013 at 12:23am

Hello,

I have this function

int *find(int *arr, int numElemns, int *value) 
{
    
    for (int i =0; i < numElemns; i++) // function to find an element in an array using linear search
    {
        if (arr[i] == *value)
        {
            return &arr[i];
        }
     }
     return NULL;
       
}

I'm kind of new to pointers and it's a bit confusing. If I wanted to return a pointer to value, how would I do it?

would the definition be like this?
int *find(int *arr, int numElemns, int *value)
How would I go from there?

I appreciate any feedback.

Last edited on Mar 20, 2013 at 1:54am

Mar 20, 2013 at 12:54am

AbstractionAnon (6954)

Yes, that definition would work.

At line 8, either return arr+i; or return &arr[i];

Note: You probably want to move line 12 to after line 14, otherwise you're going to exit the loop on the first comparison, regardless of the number of elements.

The convention for the not found condition with pointers is: return NULL;

Mar 20, 2013 at 1:57am

ZFlar (9)

I edited the original post just like you said. This is how I'm calling it:

cout << "Please enter an integer to search for: ";
            cin >> integer;
            cout << endl;
            find(numbers, arraysize, &integer);

numbers is a dynamic array. It still shows nothing. Thanks a lot for the response.

Last edited on Mar 20, 2013 at 1:57am

Mar 20, 2013 at 3:27pm

ZFlar (9)

any help :)?

Mar 20, 2013 at 5:52pm

Warnis (230)

Should work just fine, what exactly is the problem?

EDIT: Just going to ask, have you tried printing the return value? (I would say to dereference it first but you'd need to check if it's null or not before that to avoid exceptions).

Last edited on Mar 20, 2013 at 9:04pm

Mar 21, 2013 at 7:05pm

ZFlar (9)

here's my code:

int *createArray(int &size) // function to create a dynamic array of random elements
{
    int *array1;
    array1 = new int[size];
    srand(time(0));
    
    for (int i = 0; i < size; i++)
    {
        array1[i] = rand() % 20;
    }
    return array1;
}
int *find(int *arr, int size, int *value) 
{
    
    for (int i =0; i < size; i++) // function to find an element in an array using linear search
    {
        if (arr[i] == *value)
        {
            return &arr[i];
        }
    }
    return NULL;
    
}

int main()
{
    int *numbers;
        int arraysize = 0;
        int integer = 0;
        cout << "please enter an array size: ";
        cin >> arraysize;
        cout << endl << endl;

        numbers = createArray(arraysize);
        cout << "The elements of the array are: ";
        for (int i = 0; i < arraysize; i++)
        {
            cout << numbers[i] << " ";
        }
        cout << endl << endl;
        cout << "enter an integer: ";
        cin >> integer;
        cout << endl << endl;
        find(numbers, arraysize, &integer);

        system("pause");
        return 0;
}

When I call the function find, it displays nothing. Thanks again.

Mar 21, 2013 at 10:06pm

Warnis (230)

find doesn't display anything because that's not what it's doing.
Find returns a pointer to an array element, which you do nothing with.
Try

1
2
3

int *found = find(numbers, arraysize, &integer);
if(found)
    cout << *found << endl;

Last edited on Mar 21, 2013 at 10:27pm

Mar 21, 2013 at 11:07pm

ZFlar (9)

Great. Thanks a lot, Warnis.

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