Function to delete an element from an array not working - c++

Question

I wanted to write a function which upon being called deletes an element from an array given that the parameters passed in the deleteArray function were the array, its length and the value of the element to be deleted.
Tried breaking out of the for loop while transversing through the array if the element was found and then tried using i's value in another for loop to replace the current elements with their next element.
like array[j] = array[j + 1]
Here is the code:
#include <iostream>
using namespace std;
void deleteElement(int[], int, int);
int main() {
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
cout << "\nIn main function\n";
for (int i = 0; i < length; i++) {
cout << array1[i];
}
return 0;
}
void deleteElement(int array2[], int length, int element) {
int i = 0;
for (int i; i < length; i++) {
if (array2[i] == element) {
for (int j = i; j < length; j++) {
array2[j] = array2[j + 1];
}
break;
}
}
if (i == (length - 1)) {
cout << ("Element doesn't exist\n");
}
cout << "Testing OP in deleteElement\n";
for (int i = 0; i < length; i++) {
cout << array2[i];
}
}
Expected:
Testing OP in deleteElement
14356
In main function
1356
Actual:
Testing OP in deleteElement
14356
In main function
14356

Answer 1

The problem is rather silly:
At the beginning of deleteElement(), you define i with int i = 0;, but you redefine another variable i as a local index in each for loop. The for loop introduces a new scope, so the int i definition in the first clause of the for loop defines a new i, that shadows the variable with the same name defined in an outer scope.
for (int i; i < length; i++) {
And you do not initialize this new i variable.
There are 2 consequences:
undefined behavior in the first loop as i is uninitialized. The comparison i < length might fail right away.
the test if (i == (length - 1)) { tests the outer i variable, not the one that for iterated on. Furthermore, the test should be if (i == length) {
There are other issues:
the nested for loop iterates once too many times: when j == length - 1, accessing array[j + 1] has undefined behavior.
you do not update length, so the last element of the array is duplicated. You must pass length by reference so it is updated in the caller's scope.
Here is a corrected version:
#include <iostream>
using namespace std;
void deleteElement(int array2[], int& length, int element);
int main() {
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, &length, 4);
cout << "\nIn main function\n";
for (int i = 0; i < length; i++) {
cout << array1[i] << " ";
}
return 0;
}
void deleteElement(int array2[], int& length, int element) {
int i;
for (i = 0; i < length; i++) {
if (array2[i] == element)
break;
}
if (i == length) {
cout << "Element doesn't exist\n";
} else {
length -= 1;
for (; i < length; i++) {
array2[i] = array2[i + 1];
}
}
cout << "Testing OP in deleteElement\n";
for (i = 0; i < length; i++) {
cout << array2[i] << " ";
}
}

Answer 2

If you use the algorithm function std::remove, you can accomplish this in one or two lines of code without writing any loops whatsoever.
#include <algorithm>
#include <iostream>
void deleteElement(int array2[], int& length, int element)
{
int *ptr = std::remove(array2, array2 + length, element);
length = std::distance(array2, ptr);
}
int main()
{
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
for (int i = 0; i < length; ++i)
std::cout << array1[i];
}
Output:
1356
Note that we could have written the deleteElement function in a single line:
void deleteElement(int array2[], int& length, int element)
{
length = std::distance(array2, std::remove(array2, array2 + length, element));
}
Basically, std::remove moves the removed element to the end of the sequence, and returns a pointer to the beginning of the removed elements.
Thus to get the distance from the beginning of the array to where the removed elements are located, usage of std::distance is done to give us our new length.
To remove only the first found element, std::find can be used, and then std::copy over the elements, essentially wiping out the item:
void deleteElement(int array2[], int& length, int element)
{
int *ptr = std::find(array2, array2 + length, element);
if ( ptr != array2 + length )
{
std::copy(ptr+1,array2 + length, ptr);
--length;
}
}
int main()
{
int array1[] = { 1, 4, 3, 5, 4, 6, 9 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
for (int i = 0; i < length; ++i)
std::cout << array1[i];
}
Output:
135469

Answer 3

There is no need for multiple loops in deleteElement. Additionally, your removal will fail to remove all elements (e.g. 4 in your example) if your array contains more than one 4, e.g.
int array1[] = { 1, 4, 3, 4, 5 };
You can simplify your deleteElement function and handle removing multiple occurrences of element simply by keeping a count of the number of times the element is found and by using your counter as a flag to control removal, e.g.:
void deleteElement(int array2[], int& length, int element)
{
int found = 0; /* flag indicating no. element found */
for (int i = 0; i < length; i++) { /* iterate over each element */
if (array2[i] == element) { /* check if matches current */
found += 1; /* increment number found */
continue; /* get next element */
}
if (found) /* if matching element found */
array2[i-found] = array2[i]; /* overwrite elements to end */
}
length -= found; /* update length based on no. found & removed */
}
Updating your example main() to show both pre-delete and post-delete, you could do something like the following:
int main (void) {
int array1[] = { 1, 4, 3, 4, 5 };
int length = sizeof array1 / sizeof *array1; //For length of array
cout << "\nBefore Delete\n";
for (int i = 0; i < length; i++)
cout << " " << array1[i];
cout << '\n';
deleteElement(array1, length, 4);
cout << "\nAfter Delete\n";
for (int i = 0; i < length; i++)
cout << " " << array1[i];
cout << '\n';
}
Example Use/Output
Which in the case where you array contains 1, 4, 3, 4, 5 would result in:
$ ./bin/array_del_elem
Before Delete
1 4 3 4 5
After Delete
1 3 5
While you are using an array of type int (of which there are many in both legacy and current code), for new code you should make use of the containers library (e.g. array or vector, etc...) which provide built in member functions to .erase() elements without you having to reinvent the wheel.
Look things over and let me know if you have further questions.

Answer 4

This is because the length of the array is never updated after deleting. Logically the length should decrease by 1 if the element was deleted.
To fix this, either
Pass the length by reference and decrease it by 1 if the element is actually deleted. OR
Return from the deleteElement some value which indicates that the element was deleted. And based of that, decrease the value of length in the main function.
Recalculating the array length will not help because the element is not actually deleted in memory. So the memory allocated to he array remains same.
Other issues:
The first for loop in deleteElement should run till j < length - 1.
The for loop creates a local variable i, which shadows the i variable in outer scope, so the outer i is never updated and always remains = 0