Python Coding challenge - Day 510| What is the output of the following Python Code?

 

Code Explanation:

Line 1: Function Definition

def tricky(val, s=set()):

Defines a function tricky with two parameters:

val: the value to add.

s: a set with a default argument set() (an empty set).

So s=set() is created only once and reused in all future calls unless a different s is passed.

Line 2: Add to the Set

    s.add(val)

Adds the given val to the set s.

Because s is reused, it remembers previous values added in earlier calls.

Line 3: Return the Set

    return s

Returns the modified set s.

Line 4: First Call

print(tricky(1))

tricky(1) is called.

Since no second argument is given, it uses the default set.

1 is added to the set.

The set becomes: {1}

Output: {1}

 Line 5: Second Call

print(tricky(2))

tricky(2) is called.

Uses the same default set as before (shared between calls).

Adds 2 to that set.

Now the set is: {1, 2}

Output: {1, 2}

Line 6: Third Call

print(tricky(1))

tricky(1) is called again.

Still using the same set {1, 2}.

1 is already in the set, so adding it again has no effect.

Set remains: {1, 2}

Output: {1, 2}

Final Output

{1}

{1, 2}

{1, 2}