Deleting a number from an array that is being pointed to? - c++

Question

I'm having a hard time with this little project of mine; can someone help me out? I am trying to take a pointer to an array, a number, and remove the number if it is present in the array. If the number is removed, the array shrinks. Please note at the time of running removeNumber size = 6.
main
{
int size = 5; // setting array size
int *a = new int[size]; // allocating dynamic array
// initializing array
a[0] = 0; a[1] = 10; a[2] = 20; a[3] = 30; a[4] = 40;
removeNumber(a, 10, size);
}
And now the prototype:
void removeNumber(int *& arrayPtr, int number, int &size)
{
int index = check(arrayPtr, number, size);
int *newArray = new int[size - 1];
if (index != -1)
{
for (int i = number; i <= size; ++i)
newArray[i] = *&arrayPtr[i + 1];
delete[] arrayPtr;
arrayPtr = newArray;
size -= 1;
}
}
Check for reference:
int check(int *arrayPtr, int number, int size)
{
for (int i = 0; i < size; i++)
{
if (arrayPtr[i] == number) return i;
}
return -1;
}

Answer 1

The function is wrong. First of all it has a memory leak in case when the target value is not found in the array.
void removeNumber(int *& arrayPtr, int number, int &size)
{
int index = check(arrayPtr, number, size);
int *newArray = new int[size - 1];
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
if (index != -1)
{
//...
}
}
Secondly there is used incorrect range of indices
for (int i = number; i <= size; ++i)
^^^^^^ ^^^^^^^^^
newArray[i] = *&arrayPtr[i + 1];
^^^^^^
And you could write this expression *&arrayPtr[i + 1] simpler just like arrayPtr[i + 1].
You forgot to copy the elements of the original array before the index index. number is not a valid index.
Also it would be better if the function returned a Boolean value saying whether the operation was successful or not. You can change the size of the array in the calling function if the call was successful.
So I would write the function like
bool removeNumber( int * &arrayPtr, int n, int value )
{
int index = check( arrayPtr, number, size );
if ( index != -1 )
{
int *tmp = new int[size - 1];
int i = 0;
for ( ; i < index; ++i ) tmp[i] = arrayPtr[i];
for ( ; i < size - 1; i++ ) tmp[i] = arrayPtr[i+1];
delete [] arrayPtr;
arrayPtr = tmp;
}
return index != -1;
}

Answer 2

It is ok to reinvent the wheel in order to learn. However you should be aware of the standard way to handle this: using std::vector along with std::remove/std::remove_if (defined in #include <algorithm>) would make your life much simpler, with the benefit of a performant implementation.
void removeNumber(std::vector<int>& vec, int number) {
vec.erase(std::remove(std::begin(vec), std::end(vec), number),
std::end(vec));
// If you really want to shrink the vector, you can call...
vec.shrink_to_fit();
}
I also made a little demo to show you that it works as intended.

Answer 3

Your problem is here:
for (int i = number; i <= size; ++i)
First of all you start the iteration from the value of the number that you searched(in your case 10) and as a result you will not enter the for loop.So the correct starting value of i is the first number of the array which is 0.Then this line of code here:
newArray[i] = *&arrayPtr[i + 1];
doesn't even make sense(as a thought process not syntactically), but what i think you want to do here is just copy the correct element from the old array to the new one.
A correct implementation of the previous is the following code:
void removeNumber(int *& arrayPtr, int number, int &size)
{
int index = check(arrayPtr, number, size);
if (index != -1)
{
int *newArray = new int[size - 1];
int i = 0,j=0; //i for the old array and j for the new
while(i < size)//while we are in bounds
{
if(i==index)//if the index is the index of the given element
{ //skip this number
i++;
continue;
}
newArray[j++] = arrayPtr[i++];//copy the correct number to the correct position
}
delete[] arrayPtr;
arrayPtr = newArray;
size -= 1;
}
}