Segmentation fault in merge sort code - c++

Question

I have written this merge sort program in c++ but I am getting "Segmentation fault (core dumped)" error after running the code. Even though there is no compilation error. Can you please tell me what's the mistake I am doing? While taking input in the array it is showing that error. If I change it to push_back, the input is fine but later in merge function, it is showing the same error.
//merging 2 sorted subarrays.
#include <iostream>
#include <vector>
using namespace std;
void merge(vector <int> &a,vector <int> &b,vector <int> &c)
{
int i=0,j=0,k=0,bL=b.size(),cL=c.size();
while(i<bL && j<cL)
{
if(b[i]<c[j])
{
a[k]=b[i];
i++;k++;
}
else
{
a[k]=c[j];
j++;k++;
}
}
while(i<bL)
{
a[k]=b[i];
i++;k++;
}
while(j<cL)
{
a[k]=c[j];
j++;k++;
}
cout<<"array a inside merge is: "<<endl;
for(int p=0;p<a.size();p++)
{
cout<<a[p]<<endl;
}
}
void mergeSort(vector <int> &a)
{
vector <int> l, r;
int mid;
if(a.size()<2) return;
mid = a.size()/2;
for(int i=0;i<mid;i++)
{
l[i]=a[i];
}
for(int i=mid;i<a.size();i++)
{
r[i-mid]=a[i];
}
mergeSort(l);
mergeSort(r);
merge(a, l, r);
}
int main()
{
int n;
vector <int> a;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
}
mergeSort(a);
for(int p=0;p<n;p++)
{
cout<<a[p]<<endl;
}
return 0;
}

Answer 1

Whever you access an element in a vector using [] you might get a seg fault.
This code,
vector <int> a;
gives you an empty vector.
Asking for a[0] won't work if there's nothing in a yet.
Trying to set a[0] to a value won't work either.
It doesn't exist.
Yet.
You have a similar problem in mergeSort when you use
vector <int> l, r;
These are also empty vectors.
You can use push_back (or even emplace_back) to add new elements.
Or use a constructor overload to state how many elements you want.
For example,
vector <int> a(10);
gives you a vector of ten ints, so a[0] is fine to read from or write to.
a[11] isn't.
Practise using a vector first then try your merge sort.

Answer 2

The reason for getting a segmentation fault is you access the location of the memory which does not exist (more accurate to say as not allocated). Say you have a vector of length 3 and you try to access 4th position, you get a segmentation fault.
As opposed to #doctorlove's answer, I would say it is possible to use []. However, you need the following implementation (only shown for main(), please implement with the same logic in the other functions). See the documentation of std::vector for more info.
int main()
{
size_t n;
std::cin >> n;
std::vector <int> a(n);
for(int i=0;i<n;++i)
{
std::cin >> a[i];
}
// mergeSort(a);
for(int i=0;i<n;++i)
{
std::cout << a[i] << "\n";
}
return 0;
}
Hope this helps. Cheers.

Answer 3

here is the final code after changes:
//merging 2 sorted subarrays.
#include <iostream>
#include <vector>
using namespace std;
void merge(vector <int> &a,vector <int> &b,vector <int> &c)
{
int i=0,j=0,k=0,bL=b.size(),cL=c.size();
while(i<bL && j<cL)
{
if(b[i]<c[j])
{
a[k]=b[i];
i++;k++;
}
else
{
a[k]=c[j];
j++;k++;
}
}
while(i<bL)
{
a[k]=b[i];
i++;k++;
}
while(j<cL)
{
a[k]=c[j];
j++;k++;
}
cout<<"array a inside merge is: "<<endl;
for(int p=0;p<a.size();p++)
{
cout<<a[p]<<endl;
}
}
void mergeSort(vector <int> &a)
{
vector <int> l, r;
int mid;
if(a.size()<2) return;
mid = a.size()/2;
for(int i=0;i<mid;i++)
{
l.push_back(a[i]);
}
//change2
for(int i=0;i<a.size()-mid;i++)
{
r.push_back(a[mid+i]);
}
mergeSort(l);
mergeSort(r);
merge(a, l, r);
}
int main()
{
int n;
cin>>n;
//change1
vector <int> a(n);
for(int i=0;i<n;i++)
{
cin>>a[i];
}
mergeSort(a);
cout<<"Final array is:"<<endl;
for(int p=0;p<n;p++)
{
cout<<a[p]<<endl;
}
return 0;
}