I'm trying to check if a vector is sorted or not with a for loop but it's not working properly

until I make a small change but the logic seem to be the same to me.

this is the code:

vector <int> is_list_sorted(vector <int> list_of_numbers){
   bool list_is_sorted = true; 
   for(int i = 0; i < list_of_numbers.size(); i  ){
       if(list_of_number[i] > list_of_numbers[i   1]){
           list_is_sorted = false; 
       }
   }
   
   return list_is_sorted; 
}

int main(){
   if(list_is_sorted(vector1){
      cout << "the list is sorted" << endl; 
   }
   else{
      cout << "the list is not sorted << endl; 
   }
  return 0;
}

I tested this function on three different vectors and the output is always the same "the list is

not sorted" until I make a small change in the actual function and instead of:

 for(int i = 0; i < list_of_numbers.size(); i  ){
       if(list_of_number[i] > list_of_numbers[i   1]){
           list_is_sorted = false; 
       }
   }

I make it:

 for(int i = 1; i < list_of_numbers.size(); i  ){
       if(list_of_number[i] < list_of_numbers[i - 1]){
           list_is_sorted = false; 
       }
   }

what's the bug?

CodePudding user response：

With the loop

for(int i = 0; i < list_of_numbers.size(); i  )

then list_of_numbers[i 1] will be out of bounds for the last element.

If you go out of bounds you will have undefined behavior.

A different solution would be to change the loop to:

for(int i = 0; i < list_of_numbers.size() - 1; i  )

Then i 1 will not be an out of bounds index anymore.

On another note: As soon as you notice that the vector isn't sorted you can return false immediately. You don't need to check the remainder of the vector.