Deleting array pointers

Question

SO,

I am running into an error which I cannot figure out (unless my understanding is just incorrect).

I have the following code:

    int doubleSize=size*2;
    int *newArr = new int[doubleSize];
    for(int i=0; i<size; i ++) {
        newArr[i]=jon[i];
    }
    size*=2;

    display(newArr);
    jon=newArr;
    display(jon);
    delete[] newArr;
    display(jon);

After the first and second calls I get exactly what I want/expect. On the third display call the 0 and 1 indices are memory addresses, the rest of the values in the indices match the previous 2 calls. What could be causing this?

I also have another follow up question, with code as I have it, will not deleting jon[] cause the 'old' jon[] to stay in memory?

Thanks!

Answer 1

You have undefined behavior, so anything can happen.

int *newArr = new int[size*2];
// now newArr is a pointer to a memory area
jon=newArr;
// now jon is a pointer to the same area, whatever jon pointed to before is leaked
delete[] newArr;
// now the memory area no longer exists
display(jon);
// using jon is now illegal, it has the address of a deleted memory area

Probably the right solution is:

int *newArr = new int[doubleSize]; // allocate a new area
for( int i=0; i<size; ++i ) {       // fill it in
    newArr[i] = jon[i];
}
delete [] jon; // get rid of the old area we don't need anymore
jon = newArr;  // now jon is linked to our brand new data in a brand new area

Answer 2

When you delete[] newArr , you're unallocating the memory at address newArr . Since jon is also pointing to that same memory (because you set jon = newArr ), the memory is being overwritten with some other values (probably in your display function). What you need to do, is use some copy function (like memcpy ) to copy the data to a newly allocated block at jon , not just point jon at the same spot as newArr .

Answer 3

The last display call is trying to display jon that is the same as newArr - what you have just deleted! Thus the behaviour will be undefined.