Analysis of Java Programming Example Questions

Question 1:

<code>public static void demo01() {<br>    Integer f1 = 100, f2 = 100, f3 = 200, f4 = 200;<br>    System.out.println(f1 == f2);<br>    System.out.println(f3 == f4);<br>}</code>

Question 2:

<code>private static Integer i;<br>public static void demo02() {<br>    if (i == 0) {<br>        System.out.println("A");<br>    } else {<br>        System.out.println("B");<br>    }<br>}</code>

Question 1 answer:

true

false

Answer to question 2:

NullPointerException

Analysis:

Question 1:

The following is the source code of "autoboxing" in the Integer class:

<code>public static Integer valueOf(int i) {<br>    if (i >= IntegerCache.low && i <= IntegerCache.high)<br>        return IntegerCache.cache[i + (-IntegerCache.low)];<br>    return new Integer(i);<br>}<br></code>

The value of IntegerCache.low is -128, and the value of IntegerCache.high is 127. That is to say, when Integer is automatically boxed, if the range of the integer value is judged to be between [-128,127], the value in the integer constant pool will be used directly; if it is not in this range, a new Integer() will be created . Therefore, f1 and f2 in this question are both in the range of [-128,127] and use the same value in the constant pool. However, f3 and f4 are not in the range of [-128,127], and their values ​​​​are both new, so f3 and f4 are not the same object.

Question 2:

The default value of Integer i is null. When i==0 is executed, the right side of the equal sign is a number, so in order to perform comparison operations, Integer will be automatically unboxed (that is, Integer will be converted to int type). Obviously, if you unbox null (convert null to a number), a NullPointerException will be reported.

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