Java中的整型移位操作，为什么是“只有数值右端的低5位才有用”？

Question

大家好，最近在看《Java编程思想》，在第三章“操作符”中有这么一段：

如果对char、byte或者short类型的数值进行移位处理，那么在移位进行之前，它们会被转成int类型，并且得到的结果也是一个int类型的值。只有数值右端的低5位才有用。这样可防止我们移位超过init型值所具有的位数。（译注：因为2的5次方为32，而int型值只有32位。）

之后google查到了这篇文章：http://blog.csdn.net/showershow/article/details/6959122，不过还是没懂。

虽然译者做了注解，不过我还是不明白，为什么是“只有数值右端的低5位才有用”？有谁能够解释一下吗？谢谢！

PS. 第一次来这里，发现编辑器竟然支持Markdown，非常不错！

Answer 1

When I saw your question, I didn’t understand it either. So I wrote a simple program to see what bytecode is used for the displacement operation:

public class Test {
    public static void main(String[] args) {
        byte b = 32;
        int i = b << 4;
        System.out.println(i);
    }
}

Use javap to view the compiled bytecode:

  public static void main(java.lang.String[]);
    Code:
       0: bipush        32
       2: istore_1
       3: iload_1
       4: iconst_4
       5: ishl
       6: istore_2
       7: getstatic     #2                  // Field java/lang/System.out:Ljava/
io/PrintStream;
      10: iload_2
      11: invokevirtual #3                  // Method java/io/PrintStream.printl
n:(I)V
      14: return
}

I found that ishl is used, and its explanation is like this: http://cs.au.dk/~mis/dOvs/jvmspec/ref-_ishl.html

Shifts value2 left by the amount indicated in the five low bits of value1

So I finally understood that the "right-hand side" in the original English text does not refer to the "right end" of a certain value. "right-hand side" is a term that should be translated as "right operand".

Only the five low-order bits of the right-hand side will be used.

This sentence can be understood like this: the displacement operator only uses the lower 5 bits of its right operand.

When I saw this sentence, I understood that only the lower 5 bits of the left operand were used. Maybe you also understood it the same way.

P.S. Turn to wiki: http://en.wikipedia.org/wiki/Bitwise_operation#Shifts_in_Java

only the five lowest-order bits of the right-hand operand are used as the shift distance

The use of "right-hand operand" here more clearly indicates the right-hand operand.

Answer 2

The questioner may have misunderstood.

a = b << c  

The specification says Only the lower 5 digits on the right end of the value are useful It means c.

Let’s look at b first. During the calculation process, b is converted into int because the int type is 32 bits, that is, the value of b can be moved up to 31 bits.

If b is moved by 33 bits, only the last 5 bits are valid, so:

(For the convenience of testing, the following code is js code)

50 << 33
// output：100

is equivalent to

50 << (33%32)
// output：100

That is

50 << 1
// output：100

For the sake of rigor, the subject tested it in java by himself.

Answer 3

Because int only has 32 bits, << 32 overflows. So it is limited to the rightmost 5 digits to prevent overflow ( 2 ** 5 = 32 ).

Answer 4

My understanding is this, Java programming philosophy says>>>The value on the right side of this unsigned right shift operator cannot be greater than 31. For example, the right end of the value in int i = -1,i >>> 33 refers to 33 this value. That is, take the low-order 1 of 33 [0 0001], so only move one bit to the right