由负数转变逐位移位运算 [英] Bit wise shift operator with shift by negative number

问题描述

我读到一篇有趣的场景，当按位移位运算符的工作。如果第二个操作数是负的，请问按位移位运算的作品？ 。

即一个与所述;＆下; B，＆所述;＆下;移动通过在b比特的比特模式到左边。但是，如果b neagtive，应该不是成为一个运行时错误？

我能够到C成功运行低于$ C $，但我不明白它是如何工作的？

 公共静态无效bitwiseleftShift（CHAR testChar）
{
    INT VAL = testChar-A;
    INT结果= 1＆LT;＆LT; VAL;
    的System.out.println（逐位的1 VAL转变=+ VAL +为+结果）;
}
 

输入

  bitwiseleftShift（'A'）; // ASCII 65
   bitwiseleftShift（'0'）; // ASCII 48
 

结果

 位的1 VAL明智的转变= -32为1
   逐位的1 VAL转变= -49 32768
 

ASCII为'一'是97有人可以帮助我了解如何工作的？

解决方案

  

但如果b为neagtive，应该不是在运行时错误？

不根据Java语言规范的部分15.19 ：

  

如果左手操作数的提升的类型为int，仅右手操作数的五个最低阶位被用作移位距离。这是因为如果右手操作体进行按位逻辑与运算放大器＆; （§15.22.1）与屏蔽值0x1F的（0b11111）。因此实际所用的偏移距离总是在0至31范围，包括

所以-32的转变实际上最终为0的转变，以及-49的转变实际上为15移位结束 - 因此你看到的结果。

。

I came across an interesting scenario, When working with bitwise shift operator. If the second operand is negative, how does the bitwise shift operation works? .

i.e a << b , "<<" shifts a bit pattern to the left by b bits in a. But if b is neagtive, shouldn't it be an error at runtime ?

I am able to run the below code successfully but I don't understand how it works?

 public static void bitwiseleftShift(char testChar)
{
    int val=testChar-'a';
    int result= 1<<val;
    System.out.println("bit wise shift of 1 with val="+val+" is "+result);
}

Input

   bitwiseleftShift('A');// ASCII 65
   bitwiseleftShift('0'); // ASCII 48 

Results

   bit wise shift of 1 with val=-32 is 1
   bit wise shift of 1 with val=-49 is 32768

ASCII for 'a' is 97. Can someone help me understand how this works?

解决方案

But if b is neagtive, shouldn't it be an error at runtime?

Not according to the Java Language Specification, section 15.19:

If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

So a shift of -32 actually ends up as a shift of 0, and a shift of -49 actually ends up as a shift of 15 - hence the results you saw.

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