Why does `j = j ;` in Java leave `j` unchanged?

Post-Increment Operator in Java

The post-increment operator ( ) is a unary operator that increments a variable's value by 1. However, its semantics can be somewhat confusing.

Consider the following code segment:

<code class="java">int j = 0;
for (int i = 0; i <p>Here, the value of j remains 0 after the loop. This behavior is counterintuitive at first glance, but it can be understood by examining the steps taken by the compiler:</p>
<ol>
<li>
<strong>Evaluate the right-hand side</strong>: The expression j   is evaluated, which increments the value of j from 0 to 1 but returns its original value (0).</li>
<li>
<strong>Assign the result</strong>: The resulting value (0) is assigned to j.</li>
</ol>
<p>Essentially, the post-increment operator stores the original value of the variable before incrementing it. This is similar to the following code block:</p>
<pre class="brush:php;toolbar:false"><code class="java">int temp = j;
j++;
j = temp;</code>

In contrast, consider the following code segment:

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

Here, a is assigned the original value of b (0), and then b is incremented. This behavior adheres to the rule that assignments with post-fix increment operator evaluate the right-hand side first, increment the variable, and then assign the original value.

Applying the same logic to the earlier example, we can see why j = j results in 0:

1. Evaluate the right-hand side: j is evaluated, returning the original value of j (0).
2. Assign the result: The resulting value (0) is assigned to j, overwriting its incremented value (1).

Thus, the post-increment operator can lead to unexpected results if not properly understood. It is important to remember that it increments the variable after evaluating the expression and assigns the original value to the variable.

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