Java的二进制文字 - 对于字节值-128 [英] Java binary literals - Value -128 for byte

问题描述

由于SE 7 Java允许指定值作为二进制文字。文档告诉我，字节是一个可以容纳8位的信息，值-128到127型。

Since SE 7 Java allows to specify values as binary literal. The documentation tells me 'byte' is a type that can hold 8 Bit of information, the values -128 to 127.

现在我不知道为什么，但如果我尝试分配一个二进制字面值在Java字节我无法定义8位，但只有7如下：

Now i dont know why but i cannot define 8 bits but only 7 if i try to assign a binary literal to a byte in Java as follows:

byte b = 0b000_0000;    //solves to the value 0
byte b1 = 0b000_0001;   //solves to the value 1
byte b3 = 0b000_0010;   //solves to the value 2
byte b4 = 0b000_0011;   //solves to the value 3     

等等，直到我们获得了最后几possibilitys使用这些7位：

And so on till we get to the last few possibilitys using those 7 bits:

byte b5 = 0b011_1111;   //solves to the value 63
byte b6 = 0b111_1111;   //solves to the value 127

如果我想使它负数我要补充一个领导 - 前是这样的：

If i want to make it negative numbers i have to add a leading - in front like this:

byte b7 = -0b111_1111;   //solves to the value -127

现在一半的问题我有是，我只使用7位来形容他们告诉我什么是一个8位数据类型。下半场的是，他们不似乎螺纹为二进制补码，除非使用32位的int类型，我可以定义所有的32位（标志指示位包括在内）。

Now half of the problem i have is that i use just 7 bits to describe what they tell me is a 8 bit data type. Second half is that they dont seem to be threaded as twos complement unless using a 32bit int type where i can define all of the 32 bits ("sign indicator bit" included).

现在，当我搜索如何显示在数字范围-128有人告诉我这样做没有任何进一步的解释这样的：

Now when i search on how to display the in-range number -128 i was told to do it this way without any further explanation:

byte b8 = 0b1111_1111_1111_1111_1111_1111_1000_0000;

我可以清楚地看到，在过去的8位（1000 0000）重新做三三两两present -128 compelment使用8位，还是我从来没有感到困惑越来越尝试问我的问题：

I can clearly see that the last 8 Bit (1000 0000) do represent -128 in twos compelment using 8 Bit, still i never was confused more and try to ask my questions:

• 心不是上面的32位长NR 32位（Java的）int值？


• 为什么我分配一个32位的值设置为8位（基于Java）字节类型？

或一般：为什么我要这样给它分配

任何链接/这个信息将是伟大的！
感谢您对您走上阅读本文时，以及提前任何进一步的信息。

Any links/ informations about this would be great! Thank you for the time you took to read this as well as any further information in advance.

问候
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推荐答案

根据Java规范，

<一个href=\"http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1\">http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1

的所有的你声明（B，B1，...，和B8）使用 INT 文本，即使他们将适合在一个字节。有没有在java字节的文字，你只能使用一个int来初始化一个字节。

all your declarations (b, b1,..., and b8) use int literals, even when they would fit in a byte. There's no byte literal in Java, you can only use an int to initialize a byte.

我做了一些测试，字节neg128 = -0b1000_0000; 工作正常。 0b1000_0000 为128，所以你只需要把 - 之前标志。请注意，该 1 不是一个符号位是在所有（不考虑8位字节，想想32位整数转换成字节）。所以，如果你想指定的符号位，你需要写的所有32位，因为你已经证明。

I did some tests and byte neg128 = -0b1000_0000; works fine. 0b1000_0000 is 128, so you just need to put a - sign before it. Notice that that 1 is not a sign bit at all (don't think about 8-bit bytes, think about 32-bit ints converted to bytes). So if you want to specify the sign bit you need to write all 32 bits, as you have demonstrated.

所以字节B8 = 0b1000_0000; 是错误的，就像字节B8 = 128; 是错误的（+ 128不适合在一个字节）。您还可以强制转换与转换：

So byte b8 = 0b1000_0000; is an error just like byte b8 = 128; is an error (+128 does not fit in a byte). You can also force the conversion with a cast:

字节B =（字节）0b1000_0000;
要么
字节B =（字节）128;

演员告诉你知道128不适合在一个字节和位模式将reinter preTED为-128编译器。

The cast tells the compiler that you know 128 does not fit in a byte and the bit-pattern will be reinterpreted as -128.

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