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question 44 1 point what is i printed? public class test { public stati…

Question

question 44
1 point
what is i printed?
public class test {
public static void main(string args) {
int j = 0;
int i = ++j + j * 4;

system.out.println(\what is i? \ + i);
}
}
a 1
b 0
c 6
d 5

Explanation:

Step1: Evaluate pre - increment of j

The pre - increment operator ++j increments the value of j before it is used. Initially, j = 0. After ++j, the value of j becomes 1.

Step2: Evaluate the expression for i

The expression for i is ++j + j 4. We already know that after ++j, j = 1. Now we substitute j = 1 into the expression. First, calculate the multiplication part: j 4=1 4 = 4. Then calculate the addition part: ++j (which is 1)+(j 4) (which is 4) gives 1 + 4=5? Wait, no, wait. Wait, the pre - increment ++j changes j to 1. So the expression is 1+(1 4). Wait, 1+4 = 5? Wait, no, let's re - do it. Wait, the code is int i = ++j + j 4;. Let's break it down. The order of operations: in Java, the pre - increment is done first. So ++j makes j become 1. Then we have to evaluate the expression. The operators: multiplication has higher precedence than addition. So we calculate j 4 first. Since j is now 1 (after ++j), j 4=1 4 = 4. Then we add the result of ++j (which is 1) to this product. So 1+4 = 5? Wait, no, wait. Wait, ++j is a pre - increment, so the value of ++j is 1, and then j is 1. Then j 4 is 1 * 4 = 4. Then 1+4 = 5. Wait, but let's check again. Let's take the code step by step.

Initial j = 0.

When we do ++j, j is incremented to 1, and the value of ++j is 1.

Then we have j 4. Now j is 1, so j 4=4.

Then we add the two results: 1 + 4=5. Wait, but let's see the options. Option D is 5. Wait, but let's check again. Wait, maybe I made a mistake. Wait, the expression is ++j + j 4. Let's use the order of operations. In Java, the order of operations for arithmetic operators: multiplication and division have higher precedence than addition and subtraction. Also, unary operators (like ++) are evaluated before binary operators. So the pre - increment ++j is evaluated first. So j becomes 1, and the value of ++j is 1. Then we have j 4, where j is 1, so that's 4. Then we add 1 and 4, getting 5. So i is 5.

Answer:

D. 5