blocks|key|520319|text|从您的示例输入中，您只需要：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|520320|sbyte+something+=+-128;

byte+foo+=+(byte)(+something+%2B+128);|code-block|syntax|javascript|520321|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

From you sample input you simply want:

<pre><code>sbyte something = -128;

byte foo = (byte)( something + 128);
</code></pre>

blocks|key|520348|text|new+=+old+%2B+128;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|520349|宾果:-)|unstyled|520350|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>new = old + 128;
</code></pre>

bingo :-)

blocks|key|1717808|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1717809|sbyte+signed+=+(sbyte)input;|code-block|syntax|javascript|1717810|或|1717811|int+signed+=+input+%7C+0xFFFFFF00;|1717812|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try 

<pre><code>sbyte signed = (sbyte)input;
</code></pre>

or

<pre><code>int signed = input | 0xFFFFFF00;
</code></pre>

blocks|key|520468|text|++++public+static+byte+MakeHexSigned(byte+value)
++++{
++++++++if+(value+>+255+/+2)
++++++++{
++++++++++++value+=+-1+*+(255+%2B+1)+%2B+value;
++++++++}

++++++++return+value;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|520469|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> public static byte MakeHexSigned(byte value)
 {
 if (value &gt; 255 / 2)
 {
 value = -1 * (255 + 1) + value;
 }

 return value;
 }
</code></pre>

blocks|key|1717853|text|我相信2s补码字节最好用以下方法来完成。也许不是优雅或短暂，而是清晰而明显的。我会把它作为静态方法放在我的util类中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1717854|public+static+sbyte+ConvertTo2Complement(byte+b)
{
++++if(b+<+128)
++++{
++++++++return+Convert.ToSByte(b);
++++}
++++else
++++{
++++++++int+x+=+Convert.ToInt32(b);
++++++++return+Convert.ToSByte(x+-+256);
++++}
}|code-block|syntax|javascript|1717855|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I believe that 2s complement bytes would be best done with the following. Maybe not elegant or short but clear and obvious. I would put it as a static method in one of my util classes.

<pre><code>public static sbyte ConvertTo2Complement(byte b)
{
 if(b &lt; 128)
 {
 return Convert.ToSByte(b);
 }
 else
 {
 int x = Convert.ToInt32(b);
 return Convert.ToSByte(x - 256);
 }
}
</code></pre>

blocks|key|2073577|text|如果我没有正确地指出，您的问题是如何转换输入，这实际上是一个signed-byte+(塞特)，但是该输入存储在unsigned+integer中，然后通过将它们转换为零来避免负值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2073578|要明确的是，当您使用有符号类型(如ubyte)时，框架在场景后面使用Two's+complement，所以只要转换到正确的类型，您就会使用两个的补码。|2073579|然后，一旦完成了转换，就可以使用简单的if或条件三元运算符(?：)来限制负值。|2073580|下面介绍的函数将返回值0+(或从-128返回到-1)，返回值from+0+to+127的the+same+value。|2073581|因此，如果必须使用无符号整数作为输入和输出，则可以使用以下内容：|2073582|private+static+uint+ConvertSByteToByte(uint+input)
{
++++sbyte+properDataType+=+(sbyte)input;+//128..255+will+be+taken+as+-128..-1
++++if+(properDataType+<+0)+{+return+0;+}+//when+negative+just+return+0
++++if+(input+>+255)+{+return+0;+}+//just+in+case+as+uint+can+be+greater+than+255
++++return+input;
}|code-block|syntax|javascript|2073583|或者，IMHO可以将输入和输出更改为最适合输入和输出的数据类型(字节和字节)：|2073584|private+static+byte+ConvertSByteToByte(sbyte+input)
{
++++return+input+<+0+?+(byte)0+:+(byte)input;
}|2073585|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/cs7y5x0x%2528VS.90%2529.aspx|1|http://en.wikipedia.org/wiki/%253F%253a^0|U|B|1J|G|17|2|0|0|H|5|Y|G|0|J|2|U|2|1|0|B|1|U|D|18|E|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|F|3|G|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@$9|1M|A|1N|1|1O]]|E|$]]|$1|J|3|K|5|6|7|1P|8|@$9|1Q|A|1R|B|C]|$9|1S|A|1T|B|C]|$9|1U|A|1V|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1W|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1X|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|U|3|V|5|P|7|1Z|8|@]|D|@]|E|$Q|R]]|$1|W|3|-4|5|6|7|20|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]|14|$5|Z|10|11|E|$12|15]]]]

If I undestood correctly, your problem is how to convert the input, which is really a <code>signed-byte</code> (<a href="http://msdn.microsoft.com/en-us/library/cs7y5x0x%28VS.90%29.aspx" rel="nofollow">sbyte</a>), but that input is stored in a <code>unsigned integer</code>, and then also avoid negative values by converting them to zero.

To be clear, when you use a signed type (like <code>ubyte</code>) the framework is using <code>Two's complement</code> behind the scene, so just by casting to the right type you will be using two's complement.

Then, once you have that conversion done, you could clamp the negative values with a simple <code>if</code> or a conditional ternary operator (<a href="http://en.wikipedia.org/wiki/%3F%3a" rel="nofollow">?:</a>).

The functions presented below will return <code>0</code> for values <code>from 128 to 255</code> (or from -128 to -1), and <code>the same value</code> for values <code>from 0 to 127</code>.

So, if you must use unsigned integers as input and output you could use something like this:

<pre><code>private static uint ConvertSByteToByte(uint input)
{
 sbyte properDataType = (sbyte)input; //128..255 will be taken as -128..-1
 if (properDataType &lt; 0) { return 0; } //when negative just return 0
 if (input &gt; 255) { return 0; } //just in case as uint can be greater than 255
 return input;
}
</code></pre>

Or, IMHO, you could change your input and outputs to the data types best suited to your input and output (sbyte and byte):

<pre><code>private static byte ConvertSByteToByte(sbyte input)
{
 return input &lt; 0 ? (byte)0 : (byte)input;
}
</code></pre>

blocks|key|2073641|text|int8_t+indata;+/*+-128,-127,...-1,0,1,...127+*/
uint8_t+byte+=+indata+%5E+0x80;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2073642|异或MSB，仅此而已|unstyled|2073643|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>int8_t indata; /* -128,-127,...-1,0,1,...127 */
uint8_t byte = indata ^ 0x80;
</code></pre>

xor MSB, that's all

blocks|key|1717902|text|这是我对这个问题的解决方案，对于大于8位的数字。我的例子是一个16位的值。注意:您必须检查第一个位，看看它是否为否定的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1717903|步骤：|1717904|在变量前加上“~”，将#转换为恭维。(即)Y=+~y)|ordered-list-item|1717905|将#s转换为二进制字符串|1717906|将二进制字符串拆分为字符数组|1717907|从最正确的价值开始，增加1，跟踪携带。在字符数组中存储结果。|1717908|将字符数组转换回字符串。
私有字符串TwosComplimentMath(string+value1，string+value2)+{+char[]+binary1+=value2+1.TwosComplimentMath()；char[]+binary2+=value2+2.TwosComplimentMath()；bool+carry+=+false；char[]+calcResult+=+new+char16；for+(int+i=+15；i+>=+0；{+if+(binary1i+==+binary2i)+{+if+(binary1i+==+'1')+{+if+(进位){+calcResulti+=+'1'；进位=真；}+calcResulti+=+'0'；进位=真；}{+if+(进位){+calcResulti+=+'1'；{+calcResulti+=+'0'；进位=假；}}{如果(+calcResulti+=+'0'；进位=真；}{+calcResulti+=+'1'；进位=假；}})字符串结果=新字符串(CalcResult)；返回结果；}|1717909|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|T|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|U|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|V|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|W|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|P|$]]

Here is my solution for this problem, for numbers bigger than 8-bits. My example is for a 16-bit value. Note: You will have to check the first bit, to see if it is a negative or not.

Steps:

<ol>
<li>Convert # to compliment by placing '~' before variable. (ie. y = ~y)</li>
<li>Convert #s to binary string</li>
<li>Break binary strings into character array</li>
<li>Starting with right most value, add 1 , keeping track of carries. Store result in character array.</li>
<li>Convert character array back to string.

<pre><code>private string TwosComplimentMath(string value1, string value2)
{
 char[] binary1 = value1.ToCharArray();
 char[] binary2 = value2.ToCharArray();
 bool carry = false;
 char[] calcResult = new char[16];

 for (int i = 15; i &gt;= 0; i--)
 {
 if (binary1[i] == binary2[i])
 {
 if (binary1[i] == '1')
 {
 if (carry)
 {
 calcResult[i] = '1';
 carry = true;
 }
 else
 {
 calcResult[i] = '0';
 carry = true;
 }
 }
 else
 {
 if (carry)
 {
 calcResult[i] = '1';
 carry = false;
 }
 else
 {
 calcResult[i] = '0';
 carry = false;
 }
 }
 }
 else
 {
 if (carry)
 {
 calcResult[i] = '0';
 carry = true;
 }
 else
 {
 calcResult[i] = '1';
 carry = false;
 }
 }

 }

 string result = new string(calcResult);
 return result;

}
</code></pre></li>
</ol>

blocks|key|2073709|text|所以问题是OP的问题并不是真正的二补码转换。他给一组值增加了一个偏差，以调整范围从-128.127到0..255。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2073710|要实际执行两个补码转换，只需将有符号的值键入为无符号的值，如下所示：|2073711|sbyte+test1+=+-1;
byte+test2+=+(byte)test1;|code-block|syntax|javascript|2073712|-1改为255。-128改为128。不过，这听起来不像是行动小组想要的。他只想滑动一个数组，这样最低的有符号值(-128)就变成了最低的无符号值(0)。|2073713|要添加偏差，只需执行整数加法：|2073714|newValue+=+signedValue%2B128;|2073715|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

So the problem is that the OP's problem isn't really two's complement conversion. He's adding a bias to a set of values, to adjust the range from -128..127 to 0..255. 

To actually do a two's complement conversion, you just typecast the signed value to the unsigned value, like this:

<pre><code>sbyte test1 = -1;
byte test2 = (byte)test1;
</code></pre>

-1 becomes 255. -128 becomes 128. This doesn't sound like what the OP wants, though. He just wants to slide an array up so that the lowest signed value (-128) becomes the lowest unsigned value (0).

To add a bias, you just do integer addition:

<pre><code>newValue = signedValue+128;
</code></pre>

blocks|key|1717724|text|您可以简单地描述一些事情，比如在数字中添加一个偏差(在本例中，将128添加到签名数字中)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1717725|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You could be describing something as simple as adding a bias to your number ( in this case, adding 128 to the signed number ).

I need to convert bytes in two's complement format to positive integer bytes.
The range -128 to 127 mapped to 0 to 255.

<pre><code>Examples: -128 (10000000) -&gt; 0 , 127 (01111111) -&gt; 255, etc.
</code></pre>

EDIT To clear up the confusion, the input byte is (of course) an unsigned integer in the range 0 to 255. BUT it represents a signed integer in the range -128 to 127 using two's complement format. For example, the input byte value of 128 (binary 10000000) actually represents -128.

EXTRA EDIT Alrighty, lets say we have the following byte stream 0,255,254,1,127. In two's complement format this represents 0, -1, -2, 1, 127. This I need clamping to the 0 to 255 range. For more info check out this hard to find article: <a href="http://en.wikipedia.org/wiki/Two%27s_complement" rel="noreferrer">Two's complement</a>

Two's complement conversion

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要将两个补码格式的字节转换成正整数字节。范围-128到127映射到0到255。Examples: -128 (10000000) -> 0 , 127 (01111111) -> 255, etc.编辑以消除混淆，输入字节(当然)是0到255范围内的无符号整数。但它表示的范围是一个有符号整数-128到127使用两种补码格式。例如，输入字节值128 (二进制10000000)实际上表示-128。

问二补码转换
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回答 10

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问二补码转换
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