Thread: Array of pointers not changing out of scope?

Originally Posted by LaffyTaffy

I have a PiggyBank class that stores "Coins" with the functions DropIn and GrabOut. The functions will store a pointer to the "Coins" in an growable array of pointers. (coinStorage**) The functions look like this:

DropIn is passed the array address, size, and a pointer to the coin that is supposed to be "Dropped In".

Code:

bool PiggyBank::DropIn(Coin** coinStorage, int& size, Coin* coin) {
    Coin** temp = 0;
    temp = new Coin* [size+1];
    if(!temp)
        return false;

    for(int x=0; x<size; x++)
        temp[x] = coinStorage[x];
    temp[size] = coin;
    size++;

    if(coinStorage)
        delete [] coinStorage;
    coinStorage = temp;
    return true;
}

GrabOut is passed the array address and size, makes a copy of the passed array of pointers with size-1, and returns the Coin* that was "removed" from the array.

Code:

Coin* PiggyBank::GrabOut(Coin** coinStorage, int& size) {
    if(size) {
        Coin* save = coinStorage[size-1];
        Coin** temp = new Coin* [size-1];

        for(int x=0; x<size-1; x++)
            temp[x] = coinStorage[x]; 
        size--;

        if(coinStorage)
            delete [] coinStorage;
        coinStorage = temp;
        return save;
    }
    else {
        cerr << "GrabOut(): No coins to grab!\n";
        return 0;
    }
}

My problem is that it appears as though the functions are not actually changing coinStorage "out of scope", because when I call DropIn() twice, on the second call, I assumed that the value from the first call should be in coinStorage[0], but it is not. What am I missing here?

One solution was for me to modify DropIn() to return temp** instead of a bool, and have the storage = the returned ** outside of the function.. but I'm still curious why my original method didn't work.

I do realize I could use std::vector, but this is an exercise, and we are supposed to create our own growable array.

Thanks in advance for any help!

Are you required to supply that exact interface? Because it's an incredibly ass-backwards design, IMO - it violates encapsulation, RAII, and just plain common sense. The sane approach is to simply design your own vector-type class (or some reasonable facsimile thereof), a shared-pointer class, and then move forward from there. If that isn't allowed, well, I'd say that your professor is teaching you some very bad lessons!

Anyway, the reason that doesn't work is because pointers are passed by value, so you'd need to pass either a Coin*** or a Coin**&.