QUESTION 60 Given the code fragment: int a[] = {1, 2, 3, 4, 5}; for (XXXX) { System.out.print (a[e]); } Which option can replace xxxx to enable the code to print 135?

The correct answer is B. int e=0; e<5; e+=2. Option B (int e=0; e<5; e+=2) starts at index 0 and increments by 2 each iteration, producing indices 0, 2, and 4 - which map to array values 1, 3, and 5 respectively, exactly the desired output. Why the distractors fail: A (e++) increments by 1, visiting every index (0–4) and…

Using Loop Constructs

Question

QUESTION 60
Given the code fragment:
int a[] = {1, 2, 3, 4, 5};
for (XXXX) {
System.out.print (a[e]);
}
Which option can replace xxxx to enable the code to print 135?

Options

Aint e=0; e<=4; e++
Bint e=0; e<5; e+=2
Cint e=0; e<=5; e+=1
Dint e=1; e<=5; e+=2

How the community answered

(40 responses)

A
5% (2)
B
78% (31)
C
13% (5)
D
5% (2)

Explanation

Option B (int e=0; e<5; e+=2) starts at index 0 and increments by 2 each iteration, producing indices 0, 2, and 4 - which map to array values 1, 3, and 5 respectively, exactly the desired output.

Why the distractors fail:

A (e++) increments by 1, visiting every index (0–4) and printing all five values: 1 2 3 4 5.
C (e<=5) has the same step-by-1 problem as A, and also goes out of bounds at index 5, throwing an ArrayIndexOutOfBoundsException.
D starts at e=1 (value 2, not 1), skips the first element, and also throws an ArrayIndexOutOfBoundsException when e reaches 5.

Memory tip: To print every other element starting from the first, remember the pattern: start at 0, step by 2, stop before the length - e=0; e<length; e+=2. The < (strict less-than) is your safety guard against going out of bounds.

Topics

#for loop syntax#array indexing#iteration control#loop increment

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