Question

what is y displayed? 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 3. b y is 1. c y is 2. d y is 4. question 48

Explanation:

Step1: Analyze the post - increment operator

In Java, the post - increment operator x++ uses the current value of x for the operation and then increments x by 1. Initially, x = 1. When we have the expression y = x + x++, the first x is 1, and the x++ also uses the value of x as 1 (because it is post - increment) for the addition operation. So the addition is 1+1.

Step2: Calculate the value of y

After the addition operation 1 + 1, the value of y is calculated as 2. Then, the x will be incremented to 2, but this increment does not affect the value of y which has already been calculated.

Answer:

C. y is 2.