Question

what is x after the following statements? int x = 1; x *= x + 1; a x is 4. b x is 2. c x is 3. d x is 1.

Explanation:

Step1: Analyze initial value

The initial value of \( x \) is given as \( x = 1 \).

Step2: Evaluate the expression \( x *= x + 1 \)

The operator \( *= \) is a compound assignment operator, which is equivalent to \( x = x (x + 1) \). Substituting \( x = 1 \) into the right - hand side of the equation: first, calculate \( x + 1=1 + 1 = 2 \), then calculate \( x\times(x + 1)=1\times2 = 2 \)? Wait, no, wait. Wait, the initial \( x = 1 \), when we do \( x *=x + 1 \), the operation is \( x=x\times(x + 1) \). Wait, no, let's re - calculate. The original \( x = 1 \), \( x+1=2 \), then \( x=x\times(x + 1)=1\times2 = 2 \)? But that's not correct. Wait, no, wait, maybe I made a mistake. Wait, the code is int x = 1; x =x + 1;. Let's expand the x =x + 1 statement. In Java (or C - like languages), the expression x = y is equivalent to x = x y. So here \( y=x + 1 \). So first, we calculate \( y=x + 1 \) with the original value of \( x \) (which is 1), so \( y = 1+1 = 2 \). Then we do \( x=x\times y=1\times2 = 2 \)? But that's not matching the options? Wait, no, wait, maybe I misread the options. Wait, the options are A:4, B:2, C:3, D:1. Wait, maybe I made a mistake in the expansion. Wait, no, let's check again. Wait, the statement is x =x + 1. Let's compute it step by step. The initial \( x = 1 \). The expression \( x + 1 \) is evaluated first (since the order of operations: in the compound assignment, the right - hand side is evaluated first with the current value of \( x \)). So \( x+1=1 + 1 = 2 \). Then \( x=x\times(x + 1)=1\times2 = 2 \). So the value of \( x \) is 2.

Answer:

B. x is 2.