C++ position in index removal of hard coded list of numbers

Question

I have a pretty simple question. My goal for this short program is for it to display the set of numbers(which is hard coded), then have the user specify which index a number of the array should be deleted from. It then outputs the new array. This program works but has one major error. When I run it and choose position 2 for example, which should delete 45, instead deletes 34. The program outputs :

12 45 2 8 10 16 180 182 22

instead of : 12 34 2 8 10 16 180 182 22

notice that the number position I want removed instead removes in the position before the number I actually want removed, if you remember that lists start at 0. Thank you!

//This program demos basic arrays

#include <iostream>
using namespace std;

const int CAP = 10;

int main()
{
    //read index from user, delete number in position of index they specify. 
    //when printing the list, number should be gone. 
    int size;
    int list[CAP] = { 12, 34, 45, 2, 8, 10, 16, 180, 182, 22 };
    size = 10;
    int i, delIndex;

    cout << "Your list is: " << endl;
    for (i = 0; i < CAP; i++)
    {
        cout << list[i] << endl;
    }

    cout << "\nPlease enter index to delete from: ";
    cin >> delIndex;

    for (i = delIndex; i <= 10; i++)
    {
        list[i - 1] = list[i];

    }
    cout << "The index position you specified has been deleted." << endl;
    cout << "The new array is: " << endl;
    for (i = 0; i < (size - 1); i++)
    {
        cout << list[i] << endl;
    }
    return 0;
}

Your code copies all elements from your entered index to the element left of it. Index 0 is 12, Index 1: 34, Index 2: 45. Your copy operation copies index 2 to the position 2-1 (copy 45 to 34). — rbf, Feb 27 '18 at 09:32
Do you know how I would fix this? I have been messing around with that second for statement for ages and I cant figure it out as I am new to c++... — user8930130, Feb 27 '18 at 09:34

score 1 · Accepted Answer · answered Feb 27 '18 at 09:35

1

Replace this:

for (i = delIndex; i <= 10; i++)
{
    list[i - 1] = list[i];
}

with that:

for (i = delIndex; i < size-1; i++)
{
    list[i] = list[i+1];
}

answered Feb 27 '18 at 09:35

Montassar Fatfouti

20
4

1

I think you should point out that you have fixed *two* bugs here. Firstly the obvious `i-1` vs `i` thing. Secondly, the fact that the original accessed `i[10]`, but your code doesn't. As a nit, the OP code had a variable `CAP` rather than `size`. – Martin Bonner supports Monica Feb 27 '18 at 09:55
Thanks for the comment, i thought it was obvious, but the little change i've made enables the user to delete numbers multiple times, you just need to add --size; – Montassar Fatfouti Feb 27 '18 at 15:05

score 0 · Answer 2 · answered Feb 27 '18 at 10:26

Since you are with C++, why not just using a std vector?

#include <vector>

// Initialize
int vv[10] = { 12, 34, 45, 2, 8, 10, 16, 180, 182, 22 };
std::vector<int> myvector(&vv[0], &vv[0]+10);

There are other easier ways to initialize the vector depending on the compiler. See how to initialize a vector.

Then you can just use the erase method of the vector (here I assume the user knows the indexing starts with 0, otherwise you can just put delIndex-1):

myvector.erase (myvector.begin()+delIndex);

You can easily iterate over the vector to show its contents (there are easier ways of doing this depending on the compiler, use the auto keyword).

for (std::vector<int>::iterator it = myvector.begin(); it != myvector.end(); ++it)
    std::cout << ' ' << *it;

C++ position in index removal of hard coded list of numbers

2 Answers2