为什么要以单字节开头ffffff [英] Why have leading ffffff for single byte

问题描述

我正在读取数据，有些是1字节，有些是2字节，有些是3字节，等等.对于单字节

I am reading data some are 1 byte, some are are 2 byte, some are 3 byte etc. For single byte

byte qtySat = r.readByte(); 

对于2个字节，我读为

int serialNumber = r.readUnsignedShort();

阅读一切正常，效果很好.转换为十六进制格式显示时出现的问题是有趣的情况.例如

The reading is all fine and working well. The problem when I convert to show in hex format I get funny scenarios. For example for

Integer.toHexString(qtySat )

我得到的输出为ffffffce，但我只想要ce?

I get the output as ffffffce but I just want ce ?

然后输入Integer.toHexString(serialNumber) 我得到输出，例如A1，但我希望前导00 A1导致这2字节的数据.我应该对Integer.toHexString进行哪些更改?

Then for Integer.toHexString(serialNumber) I get output as say A1 but I want the leading 00 A1 cause this 2 byte of data. What changes to Integer.toHexString should I do?

推荐答案

Integer.toHexString(qtySat )为您提供了传递给它的int值的二进制表示形式.如果传递负数byte，它将变成相同值的负数int，这说明了您看到的所有1位.

Integer.toHexString(qtySat ) gives you a binary representation of the int value passed to it. If you are passing a negative byte, it will become a negative int of the same value, which explains all the 1 bits you are seeing.

在转换为十六进制字符串之前，可以使用Integer.toHexString(qtySat & 0xff)将除低8位以外的所有其他位设置为0.

You can use Integer.toHexString(qtySat & 0xff) to set all the bits other than the low 8 bits to 0 before converting to hex String.

byte qtySat = (byte)0xce;
System.out.println (qtySat);
System.out.println (Integer.toHexString(qtySat & 0xff));

打印

-50
ce

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