Why using std::sort() ) gives me garbage values? - c++

Question

I was trying to solve a problem where I had to sort so I used the standard library std::sort function but I get a wrong output in the 2nd test case:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i;
cin>>n;
int arr[n-1];
for(i=1;i<=n-1;i++)
cin>>arr[i];
int size=sizeof(arr)/sizeof(arr[1]);
sort(arr,arr+size);
for(i=1;i<=n-1;i++)
cout<<arr[i]<<" ";
cout<<endl;
}
return 0;
}
I/P:
2
5
1 2 5 4
10
1 2 3 4 5 6 7 8 10
Expected O/P:
1 2 4 5
1 2 3 4 5 6 7 8 10
Actual O/P:
1 2 4 5
2 3 4 5 6 7 8 2013562 10

Answer 1

First of all, int arr[n-1]; is a variable length array. That's not actually part of C++, even though some compilers will tolerate it nevertheless. In most cases you can just use std::vector<int>(n-1); instead.
But look at this loop:
for(i=1;i<=n-1;i++)
cin>>arr[i];
You're starting at 1 and got all the way up to n-1, however your array goes from arr[0] to arr[n-2]. So you got undefined behavior because you're writing one past the size of the array, and you also don't write to the first position (leading to more undefined behavior when you try to sort with this uninitialized value still present).
Instead, the loop should be for(i=0;i<n-1;i++) The same applies to where you're printing it. You can then sort the vector this way:
sort(arr.begin(), arr.end());
Also note that by doing n-1 you're always reading in and handling one less value that the user inputs, I'm not sure if that's your intention. If you want that, you could also just decrease n by one after reading it in instead of writing n-1 in multiple places.