Using a function to remove duplicates from an array in C++

Question

I'm writing a program that has a user input integers into an array, calls a function that removes duplicates from that array, and then prints out the modified array. When I run it, it lets me input values into the array, but then gives me a "Segmentation fault" error message when I'm done inputing values. What am I doing wrong?

Here is my code:

#include <iostream>

using namespace std;

void rmDup(int array[], int& size)
{
        for (int i = 0; i < size; i++)
        {
            for (int j = i + 1; j < size; j++)
            {
                if (array[i] == array[j])
                {
                    array[i - 1 ] = array[i];
                    size--;
                }
            }
        }
}
int main()
{
    const int CAPACITY = 100;
    int values[CAPACITY], currentSize = 0, input;

    cout << "Please enter a series of up to 100 integers. Press 'q' to quit. ";

    while (cin >> input)
    {
       if (currentSize < CAPACITY)   
       {
           values[currentSize] = input;
           currentSize++;
       }
    }

    rmDup(values, currentSize);

    for (int k = 0; k < currentSize; k++)
    {
            cout << values[k];
    }

    return 0;
}

Thank you.

For c++ consider using std::vector<int> instead of int array[] — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Dec 10, 2013 at 18:10
Why not use the Standard Library? using namespace std; sort(begin(values), end(values)); size = unique(begin(values), end(values)) - begin(values); — dyp
– dyp, Commented Dec 10, 2013 at 18:20
I don't know those commands (sort, end, unique), but I'll definitely become familiar with them. — Carlos
– Carlos, Commented Dec 10, 2013 at 23:07
consider the boundary line case ( general technic when dealing with arrays ). When i = size - 1 , then j will become size . It mneans array[j] will be array[size] . That means crash:) — qqqqq
– qqqqq, Commented Oct 24, 2014 at 4:09

Jonas · Accepted Answer · 2017-07-05 08:13:22Z

4

for (int i = 0; i < size; i++)
{
    for (int j = i + 1; j < size; j++)
    {
        if (array[i] == array[j])
        {
            array[i - 1 ] = array[i]; /* WRONG! array[-1] = something */
            size--;
        }
    }
}

If array[0] and array[1] are equal, array[0-1] = array[0], meaning that array[-1] = array[0]. You are not supposed to access array[-1].

edited Jul 5, 2017 at 8:13

Jonas

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answered Dec 10, 2013 at 18:07

Silent Control

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1 Comment

Carlos Over a year ago

These comments got me on the right track I think. I'm amazed at how quickly this community responds. Thank you.

Sceptical Jule · Accepted Answer · 2018-02-19 19:27:53Z

1

I wouldn't make it even possible to create duplicates:

int main()
{
    const int CAPACITY = 100;

    cout << "Please enter a series of up to 100 integers. Press 'q' to quit. ";

    std::set<int> myInts;
    int input;
    while (std::cin >> input && input != 'q' && myInts.size() <= CAPACITY) //note: 113 stops the loop, too!
       myInts.insert(input);

    std::cout << "Count: " << myInts.size();
}

And do yourself a favour and don't use raw arrays. Check out the STL.

edited Feb 19, 2018 at 19:27

answered Dec 10, 2013 at 19:23

Sceptical Jule

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2 Comments

Carlos Over a year ago

I wasn't initially aware that I could accidentally create duplicate arrays. Thanks for the tip.

Sceptical Jule Over a year ago

You create not duplicate arrays but duplicate values in an array. std::set "prevents" that.

lucas92 · Accepted Answer · 2013-12-10 18:20:39Z

0

#include <algorithm>
#include <iostream>
#include <iterator>

using namespace std;
int main()
{
    vector<int> vec = {1,1,2,3,3,4,4,5,6,6};
    auto it = vec.begin();

    while(it != vec.end())
    {
        it = adjacent_find(vec.begin(),vec.end());
        if(it != vec.end())
            vec.erase(it);
            continue;
    }

    for_each(vec.begin(),vec.end(),[](const int elem){cout << elem;});
    return 0;
}

This code compiles with C++11.

answered Dec 10, 2013 at 18:20

lucas92

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5 Comments

dyp Over a year ago

I think it's faster with vec.erase(unique(begin(vec), end(vec)), end(vec));, and also that gets rid of the loop.

dyp Over a year ago

for_each(vec.begin(),vec.end(),[](const int elem){cout << elem;}); Alternatively with a range-based for: for(auto const& e : vec) std::cout<<e<<", ";

dyp Over a year ago

When you use vec.erase(it), you invalidate it. I'm not sure if it's safe to compare an invalid iterator against the end iterator (in the next loop iteration).

Sceptical Jule Over a year ago

Using std::vector for erasing is expensive bacause all subsequent elements must be re-alocated after each erase() call. std::list with constant erase time should be prefered in this case.

Carlos Over a year ago

I haven't begun using vectors yet, but will look into them. Thank you for the help.

sanjeev · Accepted Answer · 2016-08-28 15:24:48Z

0

#include<iostream>
#include<stdio.h>
using namespace std;

int arr[10];
int n;

void RemoveDuplicates(int arr[]);
void Print(int arr[]);


 int main()
{

cout<<"enter size of an array"<<endl;
cin>>n;
 cout<<"enter array elements:-"<<endl;
for(int i=0;i<n ;i++)
{
    cin>>arr[i];
}
RemoveDuplicates(arr);
Print(arr);
}

 void RemoveDuplicates(int arr[])
{
   for(int i=0;i<n;i++)
{
    for(int j=i+1;j<n;)
    {
        if(arr[i]==arr[j])
        {
            for(int k=j;k<n;k++)
            {
                arr[k]=arr[k+1];

            }
            n--;
        }
        else
            j++;
    }
}
}

void Print(int arr[])
{
   for(int i=0;i<n;i++)
   {
      cout<<arr[i]<<"  ";
    }
 }

answered Aug 28, 2016 at 15:24

sanjeev

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