Question

question 46 what is y displayed in the following code? public class test { public static void main(string args) { int x = 1; int y = x++ + x; system.out.println(\y is \ + y); } } a y is 4. b y is 2. c y is 1. d y is 3.

Explanation:

Step1: Analyze post - increment operator

In Java, the post - increment operator x++ first uses the current value of x in the expression and then increments x by 1. Initially, x = 1. When we have the expression y = x++ + x, the first x (in x++) uses the value of x as 1. After using this value, x is incremented to 2.

Step2: Calculate the value of y

Now, the second x in the expression x++ + x has the value 2 (because x was incremented after the first use). So we calculate 1+2 = 3. So y will be 3.

Answer:

D. y is 3.