C++ Changing elements of an array using only pointers

Question

I'm working on a problem where I cannot use array indexing to change elements and I'm really struggling with pointers. This is code is supposed to initialize an array with all index initialized to 0, except for indexes 0 and 1. Indexes 0 and 1 are initialized to -1. The array I get back has weird numbers in it,

int* arr(int size);
int main()
{
    int low, high;
    char again = 'y';
    high = low = 0;

    cout << "\tThe Sieve of Eratosthenes" << endl << endl;
    do
    {
        do
        {
            cout << "Enter the high boundary: ";
            cin >> high;
            cout << endl;
            if (high <= 0)
                cout << "ERROR: HIGH BOUNDARY MUST BE POSITIVE" << endl;
        } while (high < 0);

        int* thearr = arr(high);
        cout << "The prime numbers from to " << high << " are: " << endl;

        for (int ix = 0; ix <= high; ++ix)
        {
            cout << thearr[ix] << "  ";
        }

        cout << endl << endl;
        cout << endl << endl;
        cout << "Try again with new boundaries? (y/n):" << endl;
        cin >> again;

        delete[] thearr;

    } while (again == 'y');

    return 0;
}

int* arr(int size)
{
    int* thearray = new int[size];
    int last = size;
    cout << *thearray << " " << last;
    while (*thearray < last)
    {
        if (*thearray <= 1)
            thearray[*thearray] = 0;
        else
            thearray[*thearray] = -1;
        ++thearray;
        cout << *thearray;
    }
    return thearray;
}

When you allocate memory with new[], the memory will be uninitialized and have indeterminate contents. That makes the conditions in your arr function invalid and have undefined behavior. — Some programmer dude
– Some programmer dude, Commented Apr 30, 2019 at 8:06
Furthermore (also in the function arr) when the loop ends, then thearray will no longer point to the original location that new[] returned. The pointer you return will be invalid. — Some programmer dude
– Some programmer dude, Commented Apr 30, 2019 at 8:08
arr is supposed to return an initialized array and I dont know how to do that without new[] — Tayl
– Tayl, Commented Apr 30, 2019 at 8:20
First of all, *thearray is exactly equal to thearray[0]. Secondly, if the value of thearray[0] is indeterminate (and could be seen as random or garbage), then how could you use that value in a condition (like e.g. *thearray < last)? You can initialize the array, that's not a problem (std::fill or std::fill_n are very helpful here) but you can't use the contents before that initialization. — Some programmer dude
– Some programmer dude, Commented Apr 30, 2019 at 8:27

Fureeish · Accepted Answer · 2019-04-30 12:54:05Z

2

There are a few ways to initialize your array to all zeros:

Use value initialization
```
int* thearray = new int[size]();
```

Use std::fill_n

int* thearray = new int[size];
std::fill_n(thearray, size, 0);

Use std::fill

int* thearray = new int[size];
int* end = thearray + size;
std::fill(thearray, end, 0);

Use pointers and explicit loops

int* thearray = new int[size];
int* end = thearray + size;
int* begin = thearray;

while (begin < end)
{
    *begin++ = 0;
}

// After loop thearray still points to the beginning of the array

Use std::vector instead
```
std::vector<int> thearray(size);
```

If you must use raw pointers (due to the assignment or exercise conditions) then one of the first two is what I recommend.

edited Apr 30, 2019 at 12:54

Fureeish

13.5k4 gold badges39 silver badges70 bronze badges

answered Apr 30, 2019 at 8:58

Some programmer dude

411k36 gold badges420 silver badges655 bronze badges

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

Comments

Alecto · Accepted Answer · 2019-04-30 08:59:26Z

Solution 1 - just use `std::vector`

The simple way to do this is to use std::vector, because std::vector makes life easy. You don't have to delete it, and it keeps track of it's length (call .size() to get the number of elements).

We can write arr pretty simply now:

std::vector<int> arr(int size) {
    std::vector<int> vect(size); // Everything initialized to 0
    vect[0] = -1;
    vect[1] = -1;
    return vect; 
}

Solution 2 - default-initialize the array

Write now, you have

int* thearray = new int[size];

This leaves the memory uninitialized. We can initialize it to 0 just by adding a () after new int[size]:

int* thearray = new int[size](); // array initialized to 0

We can rewrite arr like this:

int* arr(int size) {
    int* thearray = new int[size]();
    thearray[0] = -1;
    thearray[1] = -1;
    return thearray;
}

Solution 3 - use pointers on an uninitialized array.

Ok ok. So maybe you're into pointers, or maybe you're a student and your professor is evil. We can use them.

int* arr(int size) {
    int* vals = new int[size]; // Create the array
    // We need to return the ORIGINAL pointer (which is vals)
    // because we need to return the original pointer, we're gonna use a 
    // a new pointer called 'scan' to modify the array
    int* scan = vals; 
    *scan = -1; // Set the value at scan to -1 (this is vals[0])
    scan += 1;  // move scan to the next value in the array
    *scan = -1; // Do it again for vals[1]
    scan += 1;

    // Set the rest of the values to 0
    for(int i = 2; i < size; i++) {
        *scan = 0; // set the value to 0
        scan += 1; // move to the next value
    }

    return vals; // Return the ORIGINAL pointer
}

Collectives™ on Stack Overflow

C++ Changing elements of an array using only pointers

2 Answers 2

Comments

Solution 1 - just use `std::vector`

Solution 2 - default-initialize the array

Solution 3 - use pointers on an uninitialized array.

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Solution 1 - just use std::vector

Solution 2 - default-initialize the array

Solution 3 - use pointers on an uninitialized array.

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related

Solution 1 - just use `std::vector`