将字节（java 数据类型）值转换为位（仅包含 8 位的字符串）
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Convert byte (java data type) value to bits (a string containing only 8 bits)
提问by davidXYZ
I need to convert a value declared as a byte data type into a string of 8 bits. Is there any Java library method that can do this? For example, it should take in -128 and output "10000000". Also, input -3 should give "11111101". (I converted these by hand.)
我需要将声明为字节数据类型的值转换为 8 位字符串。有没有可以做到这一点的Java库方法？例如，它应该输入 -128 并输出“10000000”。此外，输入 -3 应给出“11111101”。（我手工转换了这些。）
Before you assume this has been answered many times, please let me explain why I am confused.
在您认为这个问题已被多次回答之前，请让我解释一下我为什么感到困惑。
The name "byte" is a little ambiguous. So, it's been difficult following other answers. For my question, byteis referring to the java data typethat takes up 8 bits and whose value ranges from -128 to 127. I also don't mean an "array of bytes". My question is about converting a single value into its 8-bit representation only. Nothing more.
“字节”这个名字有点含糊。因此，很难遵循其他答案。对于我的问题，byte是指占用 8 位且其值范围从 -128 到 127的java 数据类型。我也不是指“字节数组”。我的问题是关于仅将单个值转换为其 8 位表示。而已。
I've tried these:
我试过这些：
byte b = -128; //i want 10000000
Integer.toBinaryString(b); //prints 11111111111111111111111110000000
Integer.toString(b, 2); //prints -10000000
If there's no built-in method, can you suggest any approach (maybe bit shifting)?
如果没有内置方法，您能否提出任何方法（可能是位移）？
回答by Evgeniy Dorofeev
Try
尝试
Integer.toBinaryString(b & 0xFF);
this gives a floating length format e.g. 4-> 100. There seems to be no standard solution to get a fixed length format, that is 4-> 00000100. Here is a one line custom solution (prepended with 0b)
这给出了一个浮动长度格式，例如4-> 100。似乎没有获得固定长度格式的标准解决方案，即4-> 00000100。这是一个单行自定义解决方案（以 开头0b）
String s ="0b" + ("0000000" + Integer.toBinaryString(0xFF & b)).replaceAll(".*(.{8})$", "");
