Understanding Java Increment Operators and JVM Stack Frame Behavior

The author analyzes a Java program that uses various increment operations to illustrate how the JVM manages local variables and the operand stack during execution.

package pers.mobian.questions01;

public class test01 {
    public static void main(String[] args) {
        int i = 1;
        i = i++;
        int j = i++;
        int k = i + ++i * i++;
        System.out.println("i="+i);
        System.out.println("j="+j);
        System.out.println("k="+k);
    }
}

Step 1: int i = 1 – a simple assignment places the value 1 in the local variable table.

Step 2: i = i++ – the post‑increment returns the original value, so i remains 1 after this statement.

Step 3: int j = i++ – the current value of i (1) is stored in j, then i is incremented to 2.

Step 4: int k = i + ++i * i++ – the expression is evaluated using the current stack state, resulting in i = 4 and k = 11.

The final values after the first program are i = 4, j = 1, and k = 11, as shown in the summary diagram.

public class test02 {
    public static void main(String[] args) {
        int i = 1;
        i = ++i;
        System.out.println(i); // result: i = 2
    }
}

For the second example, the pre‑increment i = ++i first increments i to 2, then stores the new value back, yielding i = 2.

Note: All analysis and results are specific to the Java language and its JVM execution model.