Java十六进制计算 [英] Java hex calculation

问题描述

我这样声明了long值bits:

long bits = len*8L;(304)

System.out.println(bits);输出为304

如果我使用长名称位，则得到0&分别为0.

If I use the long name bits like so I get 0 & 0 respectively.

System.out.println(bits>>(4*8));
System.out.println(0xFF&(bits>>(4*8)));

如果我使用实际数字，则分别为304和48

If I use the actual number, like so, I get 304 and 48 respectively

System.out.println(304>>(4*8));
System.out.println(0xFF&(304>>(4*8)));

我正在尝试将此Java转换为JavaScript，但是JavaScript在所有情况下都为我提供了304和48.我需要它来匹配Java并赋予0和amp;值. 0.

I'm trying to convert this Java to JavaScript but JavaScript gives me 304 and 48 in all scenarios. I need it to match the Java and give values of 0 & 0.

有什么想法吗?

编辑

接下来，为了清楚起见，我需要JavaScript等于0，以模仿Java当前的方式(等于0的两个示例在我们正在开发的程序中不会更改).

Follow up, just to be clear, I need the JavaScript equivalent to equal 0, mimicking how the Java currently does it (the two examples above that equal 0 won't be changed in what we're developing).

因此console.log(0xFF&(bits >>(4 * 8)))应该等于0，当前等于48

So console.log(0xFF&(bits>>(4*8))) should equal 0, it currently equates to 48

推荐答案

The JLS, Section 15.19 covers shifting operators in Java.

如果左侧操作数的提升类型为int，则仅将右侧操作数的最低5位用作移位距离.就像右手操作数受到了按位逻辑AND运算符&的约束. (第15.22.1节)的掩码值0x1f(0b11111).因此，实际使用的移位距离始终在0至31的范围内.

If the promoted type of the left-hand operand is int, then 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.

对于int值(例如304)，4*8或32的移位值实际上是0，因此不会发生移位.然后用0xFF进行位运算，得到48.

For an int value such as 304, the bit shift value of 4*8, or 32, is really 0, so no shifting takes place. Then a bit-and with 0xFF yields 48.

如果左操作数的提升类型为long，则仅右操作数的六个最低阶位用作移位距离.就像右手操作数受到了按位逻辑AND运算符&的约束. (§15.22.1)，其掩码值为0x3f(0b111111).因此，实际使用的移位距离始终在0至63的范围内(包括端点值).

If the promoted type of the left-hand operand is long, then only the six 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 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive.

对于long值，4*8的移位值实际上确实会向右移32位，从而产生0.

For a long value, the bit shift value of 4*8 really does shift to the right 32 bits, which yields 0.

此页面涵盖了JavaScript位移位运算符.

This page covers JavaScript bit-shift operators.

按位运算符将其操作数视为32位(零和一)的序列，而不是十进制，十六进制或八进制数字.

Bitwise operators treat their operands as a sequence of 32 bits (zeroes and ones), rather than as decimal, hexadecimal, or octal numbers.

似乎JavaScript将数字转换为32位数字，例如Java int.似乎同样的仅最低5位"规则也适用于JavaScript中的移位操作数.

It appears that JavaScript converts the number to a 32-bit number, like a Java int. It also appears that the same "only the least 5 bits" rule applies to the shift operand in JavaScript also.

console.log(304>>32);        // Don't shift!
console.log(0xFF&(304>>32)); // Don't shift!
console.log(304>>33);        // Shift by 1, not 33
console.log(0xFF&(304>>33)); // Shift by 1, not 33

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