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value x result

//why does x return 2?. isnt it suppose to be 1 because of ++x; int x = 0; for (int z = 0; z<5 ; z++) { if ((z>2)&&(++x>2)) { x+=7; } } System.out.print(x); still dont get it **result : x =2

27th Dec 2017, 12:56 AM
Gustave A C/D C ☢️ 🛸♨️🛸🛸
3 Answers
0
x is incremented twice, once when z = 3, once when z = 4.
27th Dec 2017, 1:49 AM
Alexander
Alexander - avatar
0
Right, so x starts as 0, z as 0 Loop iteration 1: z is 0, not > 2. Since the first part of && (namely z > 2) is false, we do not need to evaluate the second part (which is ++x > 2). Iteration 2: z is 1, x is still 0. z is still not > 2, so again we do nothing. Iteration 3: z 2, x 0. Still same. Iteration 4: z 3, x 0. z is finally > 2, so we evaluate the second part of &&. ++x is 1, not > 2, so false, (z > 2) && (++x > 2) = true and false = false, so we do not execute the if statement. x is incremented to 1. Iteration 5: z 4, x 1. z is > 2, so we evaluate the second part of &&. ++x is 2, not > 2, so false, and the condition is again false, so we still do not execute the if statement. x is incremented to 2. Iteration 6: z 5, x 2. Exit the loop. In the end, x is 2.
5th Jan 2018, 6:05 AM
Alexander
Alexander - avatar
- 1
You need to understand how && and ++ work. The Java Tutorial explains them well enough.
5th Jan 2018, 6:09 AM
Alexander
Alexander - avatar