How to skip a index in an array C++ - c++

Question

I am trying to write a function that loops through an array, and skips over a given index. For example.
If have the array:
int Array[20] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19};
I want the output to skip over a given index. For example index 5, would skip "4". And output an array of smaller size, by skipping the "4", or index 5.
0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
My current function to skip over an index looks like this:
void deleteArray(int original[], int origLen, int newArray[], int index){
for(int i = 0; i < origLen--; i++){
newArray[i] = original[i];
if(original[i] == index){
newArray[i] = original[i-1];
}
}
}
I'm trying to iterate through the original array and copy its indexes until it gets to a certain index, and then skip over that given index, and creating a new array. Essentially, shifting all the arrays items to the left while skipping over a given index in a new array copied from the original array.
However, in my if statement, I'm not sure how to skip over the index. My current code substitutes the index with the previous index. Outputs:
0, 1, 2, 3, 4, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
Desired output:
0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19

Answer 1

For a start, your condition is incorrect:
if(original[i] == index){
You're comparing the value of the element with the desired index, rather than the index itself (which in this case should be i).
Once past the desired index, you actually want every element in the new array to be the same as the previous element in the original array - not just one - so you need to deal with this somehow. I would suggest that you keep a boolean flag to track whether you've gone past the index, and in that case, assign original[i-1] to newArray[i]; if the flag is false, then assingn original[i] to newArray[i] (as you already do) instead.

Answer 2

You can simplify the function by using two loops instead of using one loop and performing check for every run of the loop.
void deleteArray(int original[], int origLen, int newArray[], int index){
// Get everything up to index.
// When index is 5, get everything until i = 3
for(int i = 0; i < index-1; i++) {
newArray[i] = original[i];
}
// Now get everything after index.
for(int i = index; i < origLen; i++) {
newArray[i-1] = original[i];
}
}

Answer 3

std::copy it:
auto newArrayPosition = std::copy(original, original + index, newArray);
std::copy(original + index + 1, original + origLen, newArrayPosition);

Answer 4

If you always know that your new array would be one element smaller than the original i suggest you doing like this:
int Array[20] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19};
int newArray[19];
then
void deleteArray(int original[], int origLen, int newArray[], int index){
j = 0;
for(int i = 0; i < origLen; i++){
if(i == index){
continue;
}
newArray[j] = Array[i];
j++;
}
or even you can write in the boby of your loop smth like this (if you don't want to use j):
for(int i = 0; i < origLen; i++){
if(i == index){
continue;
}
if (i < index){
newArray[i] = Array[i];
}
else {
newArray[i-1] = Array[i];
}
}

Answer 5

I would just keep two counters, one for the insert position and one for the reading position:
void deleteArray(int original[], int origLen, int newArray[], int index){
int insertPos = 0;
for(int readPos = 0; readPos < origLen; readPos++){
// If this position is equal to the index we want to skip
if (readPos == index)
{
// continue onto next loop iteration
continue;
}
// Copy over the data
newArray[insertPos] = original[readPos];
// increment the counter
insertPos++;
}
}
Like others have said, avoid i < origLen-- as that is very complicated and you actually ended up not even parsing through your whole array because you kept decrementing.