Question

What will be printed by the following code segment?

var asu =0
for (cis=5; cis<=10; cis=cis+1)
{ l if (cis \% 2==0 ) asu ∗
=cis }
document.write(asu)

Answer 1

The code segment contains a JavaScript loop that multiplies a variable by even numbers between 5 and 10. Due to an error, it will print 0 instead of the expected product of even numbers.

Step-by-step explanation:

The code provided appears to contain a JavaScript loop that starts with cis equal to 5 and increments cis by 1 until it reaches 10. Each time the loop encounters an even number, it multiplies a variable asu by that even number. However, there is a mistake as asu is initialized to 0; any multiplication with 0 will always result in 0. Because of this error in initialization, the output of the code will be 0.

Let's step through the loop:

1. For cis = 5, no action is taken because 5 is not even.
2. For cis = 6, asu should have been multiplied by 6, but since asu is 0, it remains 0.
3. This pattern continues for cis = 8 and cis = 10.

After the loop has finished, the document will write out the value of asu, which is 0.

The provided JavaScript code contains a loop that iterates through values of the variable 'cis,' starting from 5 and incrementing by 1 until reaching 10. Within the loop, there's an attempt to multiply the variable 'asu' by even values of 'cis.' However, a critical mistake is evident as 'asu' is initialized to 0. Any multiplication with 0 results in 0, rendering the entire loop ineffective.

As the loop progresses from cis = 5 to cis = 10, the conditional check for even numbers triggers multiplication with 'asu,' but since 'asu' remains 0, the outcome consistently evaluates to 0. Consequently, the final output of the code, when writing the value of 'asu,' will be 0. This error in initialization significantly impacts the intended functionality of the loop, highlighting the importance of careful variable initialization in programming.