Understanding Java Increment Operators and JVM Stack Frame Behavior

The post introduces a Java interview question series and encourages readers to follow the related WeChat public account for more Java interview problems.

It then presents a Java program ( test01 ) that manipulates variables i , j , and k using post‑ and pre‑increment operators, and prints their final values.

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);
    }
}

The analysis explains each step using the JVM stack frame concepts of the local variable table and operand stack.

1. First step

Assign int i = 1 – a simple initialization.

2. Second step

Execute i = i++ . Because the post‑increment returns the original value, i remains 1.

3. Third step

Execute int j = i++ . After this statement, the local variable table holds i = 2 while the operand stack returned the previous value 1, so j = 1 .

4. Fourth step

Execute int k = i + ++i * i++ . The evaluation order leads to i = 4 in the local variable table and k = 11 .

The resulting values are shown in the accompanying image.

5. i = ++i

A second example ( test02 ) demonstrates the pre‑increment case, where i = ++i yields i = 2 .

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

The article concludes that the reasoning and calculations are specific to the Java language and its JVM execution model.