Remove all occurrences of a number in an array in C++ [closed] - c++

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Supposed we passed in the array {1, 2, 3, 1, 2, 3, 1}, and we wanted to
remove all occurrences of the number 3. The resulting array would be
{1, 2, 1, 2, 1}, and the return value would be 5, because there are now
only 5 items in the array.
This function is wrong. I don't know how to remove the occurrences. I know my index holds the positions of the occurrences. So i need to delete those positions from my array. HELP!!
I can only use arrays, i cannot use vectors or pointers.
void test() {
int data[] = { 2, 1, 3, 2, 5, 2, 7, 8, 4, 9 };
int length = 10;
int x = 2;
int index = 0;
int counter =0;
int tempIndex = 0;
int finalLength = 0;
for (int i = 0; i < length; i++)
{
if (x != data[i]) {
tempIndex = i;
counter++;
cout << "Array: " << tempIndex << endl;
}
}
cout << "Size: " << counter << endl;
}

Answer 1

you can make use of std::remove function available in C++ you can do something like below
int data[] = { 2, 1, 3, 2, 5, 2, 7, 8, 4, 9 };
int length = 10;
int* pbegin = myints;
int* pend = data+sizeof(data)/sizeof(int);
pend = std::remove (pbegin, pend, 3);
for (int* p=pbegin; p!=pend; ++p)
std::cout << *p;

Answer 2

As others have already said until they were blue in the face (probably) you cannot resize arrays. The memory is fixed. What you can do is keep track of matching elements that you want to "remove" and shift the remaining elements to make it appear that they have been removed. The reality will be that the array in memory will still be the size you started with, but you can print the results using the computed size of your "new" array.
The following code should do what you want. The first argument is your data array, the next argument will be the fixed size of said array. The last argument is the number you want to "remove" all occurrences of. The function remove_match will return the size of your data array minus any matches it found. The array will be shifted such that only the non-matching elements will be shown if you were to print the resulting array using the return value as it's new size.
int remove_match(int array[], int length, int match)
{
size_t shift(0);
int removed(0);
for (size_t i = 0; i < length; ++i)
{
if (array[i] == match)
{
removed++;
}
else
{
array[shift++] = array[i];
}
}
return (length - removed);
}
You would invoke this function like this for example
int match = 2;
int data[] = { 2, 1, 3, 2, 5, 2, 7, 8, 4, 9 };
int length = 10;
int newLength = remove_match(data, length, match);
// validate your results
for(int i = 0; i < newLength; ++i)
{
std::cout << data[i] << " ";
}
std::cout << std::endl;

Answer 3

You can't resize a C-style array without creating a new one. It is fixed. If you are looking for counting occurrence of a particular element and tracking the sum of the new data, then you could simply do
void test() {
int data[] = { 2, 1, 3, 2, 5, 2, 7, 8, 4, 9 };
int length = 10;
int x = 2;
//int index =0;
int sum(0);
int counter = 0;
for (int i = 0; i < length; i++) {
if (x == data[i]) {
//index = i;
//--counter; //return # of ocurrences in array
//cout << "Position of ocurrences: " << index << endl;
++counter;
}else{
sum += data[i];
}
}
//for (int a = counter; a < length; a++) {
//data[a] = data[a + 1];
//cout <<"testing"<< data[a];
//}
cout << "Number of ocurences: " << counter << " sum: " << sum << endl;
}
Another approach is to assign the repeated element with zero but this will not resize the array but rather it eliminates the repeated one. This is a workaround approach but it modifies the original data.