To debug the provided search program and understand why it's producing incorrect outputs, we should consider the logic in the if
statement:
if (x != int(ar[i])):
return i
The current logic states that if the current element in the array (ar[i]
) does not equal x
, the function returns the index i
. This means that as soon as it finds an element that does not match x
, it will exit the loop and return that index, which is not the intended behavior of a search function.
Given that the output is consistently returning 0
for every number except 24
, it indicates that on the first iteration of the loop, the condition is always satisfied, causing the function to return immediately.
To identify exactly what is happening, we should print the comparison being made between x
and ar[i]
in the loop before the return statement. Therefore, the most useful variable to include in a debugging statement would be:
ar[i]
This will help us see the value of the current element being compared when it checks the condition against x
. So, adding a statement like:
print(f"Comparing: {x} with {ar[i]}")
before the return line in the loop will provide insight into why the function is returning the index prematurely.