Question

question: 9
consider the following functions.
def fn_a(num):
if num > 3:
print(\fizz\)
def fn_b(num):
if num > 2 and num < 4:
print(\fizz\)
which for loop would print
fizz
fizz
for i in range(6):
fn_b(i+1)
for i in range(6):
fn_a(i)
fn_b(i)
for i in range(6):
fn_a(i)
for i in range(6):
fn_b(i)

Explanation:

Step1: Analyze fn_a

fn_a(num) prints "Fizz" when num > 3.

Step2: Analyze fn_b

fn_b(num) prints "Fizz" when \(2 < num<4\).

Step3: Analyze each for - loop option

• Option 1: for \(i\) in range(6): fn_b(i + 1)

When \(i = 1\), \(i+1=2\), condition \(2 < num<4\) (\(num = 2\)) is False. When \(i = 2\), \(i + 1=3\), condition \(2 < num<4\) (\(num = 3\)) is False. When \(i=3\), \(i + 1 = 4\), condition \(2 < num<4\) (\(num = 4\)) is False.

• Option 2: for \(i\) in range(6): fn_a(i); fn_b(i)

For \(i = 4\), fn_a(4) prints "Fizz" (since \(4>3\)), fn_b(4) (condition \(2 < 4<4\) is False). For \(i=5\), fn_a(5) prints "Fizz" (since \(5 > 3\)), fn_b(5) (condition \(2 < 5<4\) is False).

• Option 3: for \(i\) in range(6): fn_a(i)

When \(i = 4\), fn_a(4) prints "Fizz" (since \(4>3\)). When \(i=5\), fn_a(5) prints "Fizz" (since \(5 > 3\)).

• Option 4: for \(i\) in range(6): fn_b(i)

When \(i=3\), condition \(2 < 3<4\) is True (prints "Fizz"). When \(i = 4\), condition \(2 < 4<4\) is False.

Answer:

Third option (for \(i\) in range(6): fn_a(i))