What wrong memory am I accessing which is causing segmentation fault? - c++

Question

Having Segmentation Fault on the following code of counting inversions in an array.
#include <iostream>
#include <limits.h>
using namespace std;
int merge(int *a, int p, int q, int r);
int mergesort(int *a, int p, int r) {
if (p < r) {
int q = (p + r) / 2;
//Dividing the array in two.
int l = mergesort(a, p, q);
int r = mergesort(a, q + 1, r);
int cross = merge(a, p, q, r);
return (l + r + cross);
} else
return 0;
}
This was the mergesort function supposed to divide the array. Doesn't seem to have an error here.
int merge(int *a, int p, int q, int r) {
int inv_count =0;
int n1 = q - p + 1, n2 = r - q;
int L[n1 + 1], R[n2 + 1];
//Two subarrays with last elements as inf.
L[n1] = INT_MAX; R[n2] = INT_MAX;
for (int i = p, j = 0; j < n1; i++, j++) L[j] = a[i];//copy left
for (int i = q + 1, j = 0; j < n2; i++, j++) R[j] = a[i];//copy right
int i = 0, j = 0, k = p;//merge and count
while (k <= r) {
if (L[i] <= R[j]) {
a[k++] = L[i];
i++;
} else {
a[k++] = R[j];
inv_count += n2 - j + 1;
j++;
}
}
return inv_count;
}
The problem maybe seems to have in the above function where I have declared the L and the R array.
int main() {
//code
int n; cin >> n;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
cout << mergesort(a, 0, n - 1);
return 0;
}
Example input: 5 5 4 3 2 1 shows Seg Fault.

Answer 1

In mergesort, you have a shadowing problem:
int r = mergesort(a,q+1,r);
^
The r argument passed isn't the r declared in the function, it's the r you just declared. That's undefined behavior.
By "print debugging" merge with the code:
int merge(int *a, int p, int q, int r) {
int inv_count = 0;
int n1 = q - p + 1, n2 = r - q;
std::cout << "merge(a, " << p << ", " << q << ", " << r << ')'<< std::endl;
std::cout << "\tsize of L: " << (n1 + 1) << std::endl;
std::cout << "\tsize of R: " << (n2 + 1) << std::endl;
(...)
I got the output:
merge(a, 0, 0, 0)
size of L: 2
size of R: 1
merge(a, 0, 1, 0)
size of L: 3
size of R: 0
merge(a, 0, 2, 0)
size of L: 4
size of R: -1
The segfault happens when you try to create an array with negative size. Notice the r parameter passed to merge, it's because of that first problem.

Answer 2

There are multiple problems in the code:
int r = mergesort(a, q + 1, r); redefines r and passes an uninitialized value to mergesort instead of the argument r. Use a different variable name such as:
int nr = mergesort(a, q + 1, r);
L[n1] = INT_MAX; R[n2] = INT_MAX; using sentinel values to avoid testing the index values in the merge phase is a common but nefarious example in some textbooks. The algorithm will fail if the array contains entries with the value INT_MAX, which is possible with the main() function provided. Don't use this silly approach and test the index values.
cout << mergesort(a, 0, n - 1); Passing the index to the last element of the slice leads to cumbersome code with non-intuitive + 1 and - 1 adjustments. Furthermore it does not allow passing empty slices. Passing the index of the first excluded element is more idiomatic in C and C++ and simplifies the code.
int q = (p + r) / 2; This expression can overflow and produce an incorrect value for q. It should be written: int q = p + (r - p) / 2;
index values should have type size_t.
Here is a modified version for small values of n:
#include <iostream>
using namespace std;
size_t merge(int *a, size_t p, size_t q, size_t r) {
size_t inv_count = 0;
size_t n1 = q - p, n2 = r - q;
int L[n1], R[n2];
for (size_t i = 0; i < n1; i++) L[i] = a[p + i];
for (size_t i = 0; i < n2; i++) R[i] = a[q + i];
size_t i = 0, j = 0, k = p; //merge and count
while (k <= r) {
if (i < n1 && (j == n2 || L[i] <= R[j])) {
a[k++] = L[i];
i++;
} else {
a[k++] = R[j];
inv_count += n2 - j + 1;
j++;
}
}
return inv_count;
}
size_t mergesort(int *a, size_t p, size_t r) {
if (r - p > 1) {
size_t q = p + (r - p) / 2;
// Dividing the array in two.
size_t nl = mergesort(a, p, q);
size_t nr = mergesort(a, q, r);
size_t cross = merge(a, p, q, r);
return nl + nr + cross;
} else {
return 0;
}
}
int main() {
size_t n; cin >> n;
int a[n];
for (size_t i = 0; i < n; i++) cin >> a[i];
cout << mergesort(a, 0, n);
return 0;
}