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java特性-自动装箱与解箱

March 28th, 2008 No comments

3 Autoboxing and Unboxing 自动装箱与解箱
3.1 Converting Primitives to Wrapper Types
看一下代码

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        int foo = 0;
        Integer integer = foo;

反编译class文件看到

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        int foo = 0;
        Integer integer = Integer.valueOf(foo);

是编译器帮助做了类型转换,各原始类型分别与它wrapper之间对应
Number n = 0.0f;
上面的申明显示从float转到Float,再从Float转到Number

3.2 Converting Wrapper Types to Primitives
看一下代码

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        Integer integer = 1;
        int i = integer;

反编译class文件看到

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        Integer integer = Integer.valueOf(1);
        int i = integer.intValue();

先装箱再解箱,从上代码可以看出,如果

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        Integer integer = null;
        int i = integer.intValue();

那么这个代码必定报NullPointerException,所以在解箱的时候需特别注意这点

3.3 Incrementing and Decrementing Wrapper Types
看一下代码

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        Integer counter = 1;
        while (counter<10) {
          System.out.printf("Iteration %d%n", counter++);
        }

反编译class文件看到

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        Object aobj[];
        for(Integer counter = Integer.valueOf(1); counter.intValue() < 10; System.out.printf("Iteration %d%n", aobj))
        {
            aobj = new Object[1];
            counter = Integer.valueOf(counter.intValue() + 1);
            aobj[0] = counter;
        }

在这里我们看到自增长运算符++在这里也是可以使用的,编译后实际上在在int上操作,然后再赋给counter

3.4 Boolean Versus boolean
看一下代码A

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        Boolean case1 = true;
        Boolean case2 = true;
        boolean case3 = false;
        Boolean result = (case1 || case2) && case3;

反编译class文件看到

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        Boolean case1 = Boolean.valueOf(true);
        Boolean case2 = Boolean.valueOf(true);
        boolean case3 = false;
        Boolean result = Boolean.valueOf((case1.booleanValue() || case2.booleanValue()) && case3);

在进行或运算和与运算的时候都用到了unboxing,然后再把结果autoboxing
For inquiring minds, primitives are boxed up to wrapper types in equality comparisons. For operators such as <, >=, and so forth, the wrapper types are unboxed to primitive types.
看一下代码B

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  Integer i1 = 256;
  Integer i2 = 256;
  if (i1 == i2) {
   System.out.println("Equal!");
  } else {
   System.out.println("Not equal!");
  }

在大多数的jvm里返回的是not equal,==运算符直接是对两个object intances的比较,不多内存地址,结果当然是not equal,当然也有少部分最优化的jvm可能会返回equal,如果优化的策略是创建一个instance实例
看一下代码C

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  Integer i1 = 100;
  Integer i2 = 100;
  if (i1 == i2) {
   System.out.println("Equal!");
  } else {
   System.out.println("Not equal!");
  }

返回结果是equal,记住int值在-127到127的范围里,这样的两个同int值的Integer对象jvm是只分配一个内存地址的

3.5 Conditionals and Unboxing
看一下代码A

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        Boolean arriving = false;
        Boolean late = true;
        System.out.println(arriving ? (late ? "It's about time!" : "Hello!") :
                                      (late ? "Better hurry!" : "Goodbye"));

反编译看

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        Boolean arriving = Boolean.valueOf(false);
        Boolean late = Boolean.valueOf(true);
        System.out.println(arriving.booleanValue() ? late.booleanValue() ? "It's about time!" : "Hello!" : late.booleanValue() ? "Better hurry!" : "Goodbye");

输出是Better hurry!
看一下代码B

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        String s = "hello";
        StringBuffer sb = new StringBuffer("world");
        boolean mutable = true;
        CharSequence cs = mutable ? sb : s;

反编译看到

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        String s = "hello";
        StringBuffer sb = new StringBuffer("world");
        boolean mutable = true;
        CharSequence cs = ((CharSequence) (mutable ? ((CharSequence) (sb)) : ((CharSequence) (s))));

现在的jdk中String和StringBuffer都实现了CharSequence,故String和StringBuffer都转成CharSequence

3.6 Method Overload Resolution
看一下代码

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   public static void main(String[] args) {
        int foo = 1;
        doSomething(foo);
    }
 
    public static void doSomething(double num) {
        System.out.println("double called");
    }
 
    public static void doSomething(Integer num) {
        System.out.println("Integer called");
    }

猜一下哪个被调用呢?
输入的结果是”double called”,在这里,你是不是会想到autoboxing?实际上在新的java中方法的调用遵循以下3个规则
1:不考虑boxing,unboxing及变参,而是先遵循1.4下的调用顺序
2:1.4下的规则找不到,参数boxing或者unboxing,找可调用的方法
3:以上规则找不到,考虑变参的方式调用方法

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