Python Coding Challange - Question With Answer(01050225)

 

The given code defines a function gfg that takes two arguments:

1. x: An integer that specifies how many iterations the loop will run.
2. li: A list (default is an empty list []) to which the function will append square values (i*i) based on the loop iterations.

Let’s break it down:

Code Explanation:

Function Definition:

def gfg(x, li=[]): 

• The function takes two parameters:
  • x: Number of times the loop will run.
  • li: A list that can be optionally provided. If not provided, it defaults to an empty list ([]).

Loop:

for i in range(x): 
    li.append(i*i)

• A for loop runs from i = 0 to i = x-1 (because range(x) generates values from 0 to x-1).
• For each iteration, the square of the current value of i (i*i) is calculated and appended to the list li.

Output:

print(li)

• After the loop ends, the updated list li is printed.

Function Call:

gfg(3, [3, 2, 1])

• x = 3: The loop will run 3 times (i = 0, 1, 2).
• li = [3, 2, 1]: This is the initial list provided as an argument.

---

Step-by-Step Execution:

1. Initial values:

   x = 3
   li = [3, 2, 1]

2. Loop iterations:

   • Iteration 1 (i = 0): Append 0*0 = 0 → li = [3, 2, 1, 0]
   • Iteration 2 (i = 1): Append 1*1 = 1 → li = [3, 2, 1, 0, 1]
   • Iteration 3 (i = 2): Append 2*2 = 4 → li = [3, 2, 1, 0, 1, 4]

3. Output:

   • The final list li = [3, 2, 1, 0, 1, 4] is printed.

---

Output:

[3, 2, 1, 0, 1, 4]

---

Key Notes:

1. The li=[] default argument is mutable, meaning if you don't provide a new list as an argument, changes made to li persist between function calls. This doesn't happen here because you provided a new list [3, 2, 1].

2. If you call gfg(3) without providing a list, the function will use the same default list ([]) for every call, and changes will accumulate across calls.