Java原语和包装器已升级+；拳击！JVM在这里做什么

我想买1Z0-815。我所做的代码是：

public class PromotedPlusBoxed {
    public static void main(String[] args) {
       /*AUTOMATIC PROMOTION*/   
       final byte byteValue = 10;
       final short shortValue = 10;        
       final char charValue = 10;
       final int intValue = 10;
       final long longValue = 10;
       final float floatValue = 10;
       final double doubleValue = 10;
       /**/
       final Byte byteWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (byte)10;? a performing a single boxed process.*/
       final Short shortWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (short)10;? a performing a single boxed process.*/        
       final Character charWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (char)10;? a performing a single boxed process.*/        
       final Integer intWrapperValue = 10;
       final Long longWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/
       final Float floatWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/
       final Double doubleWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/
    }    
}

我也理解原语部分，我明白我不能这样做

  final Long longWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/
  final Float floatWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/
  final Double doubleWrapperValue = 10;/*CAN'T PERFORM TWO-PHASE PROCESS 1). PROMOTED + 2).BOXED*/

因为10是一个文本int，所以2进程需要首先升级为float、long和double，然后装箱到相应的包装器中

但这是在编译

 final Byte byteWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (byte)10;? a performing a single boxed process.*/
 final Short shortWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (short)10;? a performing a single boxed process.*/        
 final Character charWrapperValue = 10;/*WHAT PROCESS IS HERE? A HIDDEN = (char)10;? a performing a single boxed process.*/    

这里的过程是什么？10被提升为字节，然后被装箱成字节，这是2进程吗？但是在这个过程中，长与字节、字符、短之间的区别是什么

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总而言之：我想理解为什么文字int可以被装箱成字节、字符、短字符而不是长字符。

答案在于字节码。如果我们想查看它，我们必须先编译它，所以将您的上一个示例更改为：

final Byte byteWrapperValue=10；
最终shortWrapperValue=10；
最终字符charWrapperValue=10；
最终整数intWrapperValue=10；
最终长包装纸值=10L；我想理解为什么一个文本int可以被装箱成字节、字符、短字符而不是长字符。仅此而已。相关：和
   L0
    BIPUSH 10
    INVOKESTATIC java/lang/Byte.valueOf (B)Ljava/lang/Byte;
   L1
    BIPUSH 10
    INVOKESTATIC java/lang/Short.valueOf (S)Ljava/lang/Short;
   L2
    BIPUSH 10
    INVOKESTATIC java/lang/Character.valueOf (C)Ljava/lang/Character;
   L3
    BIPUSH 10
    INVOKESTATIC java/lang/Integer.valueOf (I)Ljava/lang/Integer;
   L4
    LDC 10
    INVOKESTATIC java/lang/Long.valueOf (J)Ljava/lang/Long;
   L5
    LDC 10.0
    INVOKESTATIC java/lang/Float.valueOf (F)Ljava/lang/Float;
   L6
    LDC 10.0
    INVOKESTATIC java/lang/Double.valueOf (D)Ljava/lang/Double;

   L0
    LDC 10 # load the value from the constant pool
   L1
    INVOKESTATIC java/lang/Long.valueOf (J)Ljava/lang/Long; # convert the value to the wrapper type with autoboxing