Removing elements from dynamic arrays without changing the index order - c++

Question

I've been trying to erase an element from an array without changing the index order, for instance:
class MyObject
{
int id;
public:
MyObject() { }
void setId( int i ) { id = i; }
void showId() { std::cout << "ID: "<< id << "\n"; }
};
MyObject *myArray;
int main ( )
{
myArray = new myArray[6];
for( int i = 0; i < 6; i++ )
{
myArray[i]->setId(i);
myArray[i]->showId();
}
}
I want to remove myArray[3] without changing the index of the others. e.g:
myArray[0] = ID: 1
myArray[1] = ID: 2
myArray[2] = nothing
myArray[3] = ID: 4
myArray[4] = ID: 5
myArray[5] = ID: 6
I've tried to use use memset(), but it didn't work.
memset(&myArray[3],0,sizeof(MyObject));

Answer 1

There's no such thing as "nothing" in C++ language. Once you have an array, all elements of that array will contain "something". There's no way to make an array element just disappear with keeping all other elements in their original places. You can't create a hole in the array.
All you can do in this case is simply label some element as "deleted" and then later recognize it as such. The element will, of course, continue to exist physically. It is you who'll have to recognize it as "deleted" and ignore it in your further processing. You can either add some bool is_deleted field to your object, or you can use some reserved value of id (like -1) to indicate a deleted element.
In your example with memset you essentially set the id to zero. Is 0 a valid id value? If it is not, then 0 is a good choice to mark a deleted element. In that sense your memset attempt works perfectly fine, as it should. Although I'd recommend doing it by explicitly assigning zero to id, without using memset.

Answer 2

You are calling memset to write a bunch of zeros over the top of an object instance. Do not do this! You may get away with it if your class is a true POD class. You might end up just setting the ID to 0. But maybe there is more to your class that you aren't showing. In anycase, even if it isn't POD, don't use memset like that.
You can either store pointers to object and use the null pointer to indicate there is nothing there. I'd do this with std::vector<MyObject*>. Or you use a sentinel object instance, for example with ID of -1.
The other thing that could be a problem is that you appear to be using 1-based indices. C++ arrays are 0-based, so the first element is myArray[0] and not myArray[1].

Answer 3

Using memset that way is setting all the bytes of that object to 0. This is usually equivalent to setting id to 0, because the memory of an object is the memory of its members (not counting vtables, padding, etc). But don't do that anyway.
One way to do this is to use new and have an array of pointers.
MyObject* myArray[6];
int main ( )
{
for( int i = 0; i < 6; i++ )
{
myArray[i] = new MyObject;
myArray[i]->setId(i);
myArray[i]->showId();
}
}
Then to display them all:
for (int i = 0; i < 6; i++) {
cout << "myArray[" << i << "] = ";
if (myArray[i])
myArray[i]->showId();
else
cout << "nothing" << endl;
}
Then when you want to remove an object, delete it and set its pointer to NULL:
delete myArray
myArray[3] = NULL;
Then when you do anything with one of the objects in myArray, you must check if it is NULL to see if it's a valid object.

Answer 4

Consider boost::optional:
typedef boost::optional<MyObject> MyObjectOpt;
MyObjectOptArr *myArray;
The syntax/usage is a bit different (resembles using pointer):
for (int i = 0; i < 6; ++i) {
if (myArray[i])
cout << "myArray[" << i << "] = " << *(myArray[N]);
else
cout << "nothing" << endl;
}
To unset value do:
myArray[N] = boost::none;