There are several questions related to this error on stackoverflow, and I understand that its related to excess memory usage by the array, or when using pointers (I tried this with vectors aswell) but using a small array, it still shows this error. The same code earlier was running fine (for merge sorting an array).
My input was as follows:
5
9 8 1 2 4
Output:
Segmentation fault: 11
#include<iostream>
#include<vector>
using namespace std;
void merge(vector <int> ar, int l, int m, int r){
int n1 = m-l+1;
int n2 = r-m;
int L[n1];
int R[n2];
for (int i = 0; i < n1; ++i)
{
L[i]=ar[l+i];
}
for (int j = 0; j < n2; ++j)
{
R[j]=ar[m+j+1];
}
int i,j;
i = j = 0;
int k = i;
while(i<n1 && j<n2){
if (L[i]<R[j])
{
ar[k]=L[i];
i++;
}
else if (R[j]<L[i])
{
ar[k]=R[j];
j++;
}
k++;
}
while(i<n1){
ar[k]=L[i];
i++;
k++;
}
while(j<n2){
ar[k]=R[j];
j++;
k++;
}
}
void mergesort(vector <int> ar, int l, int r){
int m;
m=r+(l-r)/2;
if (l<r)
{
mergesort(ar, l, m);
mergesort(ar, m+1, r);
merge(ar, l, m, r);
}
}
void print(vector <int> ar, int size){
for (int i = 0; i < size; ++i)
{
cout<<ar[i]<< " ";
}
}
int main()
{
int n;
cin>>n;
vector <int> ar;
for (int i = 0; i < n; ++i)
{
cin>>ar[i];
}
print(ar,n);
mergesort(ar, 0, n-1);
print(ar, n);
return 0;
}
The problem is in part with m=r+(lr)/2
. When l
is 0
and r
is 1
, (lr)/2
is 0
. This makes m
equal to 1
, l
equal to 0
, and r
equal to 1
and the mergesort(ar, l, m);
call identical to the one it just worked through. The stack grows unbounded until you have a segmentation fault. One way to fix this which will also make your code more efficient is to merge the lists when the difference between l
and r
is below some threshold. Or, you can just swap the two elements when you get to the point where l
and r
differ by one, like so:
if (l - r <= 1) {
int temp = ar[l];
ar[l] = ar[r];
ar[r] = temp;
return;
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.