var i = 10; var j = 5; var z = (++i + j++) / i - 2; System.out.println(i); What is the result?

The correct answer is E. 12. There appears to be an error in the provided answer key - D (11) is actually the correct answer, not E (12). Here's the step-by-step trace: | Step | Expression | i | j | Value | |------|-----------|-----|-----|-------| | Start | - | 10 | 5 | - | | ++i | pre-increment | 11 | 5 |…

Working with Java Data Types

Question

var i = 10;
var j = 5;
var z = (++i + j++) / i - 2;
System.out.println(i);
What is the result?

Options

How the community answered

(26 responses)

A
8% (2)
B
4% (1)
D
12% (3)
E
77% (20)

Explanation

There appears to be an error in the provided answer key - D (11) is actually the correct answer, not E (12).

Here's the step-by-step trace:

Step	Expression	`i`	`j`	Value
Start	-	10	5	-
`++i`	pre-increment	11	5	11
`j++`	post-increment	11	6	5 (old value used)
`(11 + 5) / i`	`i` is now 11	11	6	`16 / 11 = 1` (integer division)
`1 - 2`		11	6	`z = -1`

System.out.println(i) prints 11 because ++i (pre-increment) raised i from 10 to 11, and nothing else modifies i after that.

Why the distractors fail:

A (5) and B (3) - these would only arise from confusing z with i, or miscounting increments
C (10) - this forgets that ++i modifies i before use (unlike i++)
E (12) - this would require i to be incremented twice, but there is only one ++i in the expression

Memory tip: The prefix ++i vs. postfix i++ rule: prefix = increment before using the value (and thus modifies i immediately), postfix = use the value first, increment after. In ++i + j++, only i is changed before the expression evaluates.

Topics

#increment operators#pre vs post increment#operator precedence#expression evaluation

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