Why Does `j = j ` Result in `j` Staying at 0 in Java?

Post Increment Operator in Java

The following code snippet from Joshua Bloch's "Java Puzzlers" seems puzzling at first glance:

<code class="java">public class Test22 {
    public static void main(String[] args) {
        int j = 0;
        for (int i = 0; i < 100; i++) {
            j = j++;
        }
        System.out.println(j); // prints 0

        int a = 0, b = 0;
        a = b++;
        System.out.println(a); // prints 0
        System.out.println(b); // prints 1
    }
}</code>

Confusion Over Post Increment

The confusing part is the behavior of j = j , which results in j remaining at 0. According to the author, this expression is similar to:

<code class="java">temp = j;
j = j + 1;
j = temp;</code>

However, in the case of a = b , b indeed becomes 1, which seems inconsistent.

Understanding the Difference

To resolve this confusion, we need to understand the rule for postfix increment operator ( ):

lhs = rhs++;

Is equivalent to:

temp = rhs;
rhs = rhs + 1;
lhs = temp;

Applying the Rule

Applying this rule to a = b , it becomes:

temp = b;
b = b + 1;
a = temp;

Hence, a gets the original value of b (which is 0), and b is then incremented to 1.

In the Case of j = j

Similarly, applying the rule to j = j yields:

temp = j;
j = j + 1;
j = temp;

Since temp is equal to the original value of j (which is 0), j is incremented to 1 but then immediately overwritten with the original value (0), resulting in j remaining at 0.

Conclusion

The key to understanding the behavior of post increment operator is to remember the rule that lhs = rhs is a two-step process: first, the current value of rhs is assigned to temp, then rhs is incremented, and finally temp is assigned to lhs.

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