blocks|key|1465678|text|怎么样|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465679|myInt+=+myInt+*+-1|code-block|syntax|javascript|1465680|​|1465681|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

How about

<pre><code>myInt = myInt * -1
</code></pre>

blocks|key|1250822|text|int+myNegInt+=+System.Math.Abs(myNumber)+*+(-1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1250823|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>int myNegInt = System.Math.Abs(myNumber) * (-1);
</code></pre>

blocks|key|1250869|text|int+negInt+=+-System.Math.Abs(myInt)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1250870|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>int negInt = -System.Math.Abs(myInt)
</code></pre>

blocks|key|1465749|text|用同样的方法使任何其他的东西都是负面的:在它前面加上一个负号。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465750|var+positive+=+6;
var+negative+=+-positive;|code-block|syntax|javascript|1465751|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|$]]

The same way you make anything else negative: put a negative sign in front of it.

<pre><code>var positive = 6;
var negative = -positive;
</code></pre>

blocks|key|1465694|text|简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465695|myInt+*=+-1;|code-block|syntax|javascript|1465696|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|$]]

The easy way:

<pre><code>myInt *= -1;
</code></pre>

blocks|key|1250787|text|int+negInt+=+0+-+myInt;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1250788|或者保证是负的。|unstyled|1250789|int+negInt+=+-System.Math.Abs(someInt);|1250790|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>int negInt = 0 - myInt;
</code></pre>

Or guaranteed to be negative.

<pre><code>int negInt = -System.Math.Abs(someInt);
</code></pre>

blocks|key|1465898|text|要切换整数的符号，只需使用符号运算符：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465899|myInt+=+-myInt;|code-block|syntax|javascript|1465900|若要使其为负值，而不管原始值是否为负值，请首先使用Abs方法：|1465901|myInt+=+-Math.Abs(myInt);|1465902|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|$]]

To switch the sign of an integer, you just use the sign operator:

<pre><code>myInt = -myInt;
</code></pre>

To make it negative regardless if the original value is negative or not, you first use the Abs method:

<pre><code>myInt = -Math.Abs(myInt);
</code></pre>

blocks|key|1466222|text|我使用myInt+=+-myInt;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1466223|如此简单和容易|1466224|如果你只想要正面的|1466225|myInt+=+myInt>0+?+myInt+:+-myInt;|code-block|syntax|javascript|1466226|entityMap^0|3|F|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|U|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|V|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I use <code>myInt = -myInt;</code>

So simple and easy

If you want to have only positive

<pre><code>myInt = myInt&gt;0 ? myInt : -myInt;
</code></pre>

blocks|key|1465825|text|编辑:这对于正输入是错误的...我犯了一个错误，忘记了-x中的其余位(2s补码值)与它们在%2Bx中的值是相反的，而不是相同的。因此，简单地更改符号位对于正数是无效的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465826|我把这个留在这里...|1465827|或者更棘手的方式(我认为)……|1465828|int+y=x%7C+~int.MaxValue；|1465829|cause+int.MaxValue+is+0111+1111+1111+1111+1111+1111+1111+1111|code-block|syntax|javascript|1465830|所以|1465831|~int.MaxValue+is++++++1000+0000+0000+0000+0000+0000+0000+0000|1465832|因此，任何与之进行或运算的int32都将在符号位中放置1(使其为负)，并使所有其他位保持不变……|1465833|编辑:实际上，由于1000+0000+0000+0000实际上是Minvalue，这也应该是可行的：|1465834|+++int+y+=+x+%7C+int.MinValue;+//+or,+to+do+it+to+itself,
+++x+%7C=+int.MinValue;|1465835|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|11|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

EDIT: This is wrong for positive inputs... I made mistake of forgetting that the rest of the bits in -x (2s-Complement value) are the 'opposite' of their value in +x, not the same. SO simply changing the sign bit will NOT work for positive numbers. 

I'll leave this here for in for purposes... 

Or the tricky way ( I think )...

int y = x | ~int.MaxValue;

<pre><code>cause int.MaxValue is 0111 1111 1111 1111 1111 1111 1111 1111
</code></pre>

so 

<pre><code>~int.MaxValue is 1000 0000 0000 0000 0000 0000 0000 0000
</code></pre>

and therefore any int32 Or'ed with that will put a 1 in the sign bit, (making it negative), and leave all the other bits the same... 

EDIT:
 actually, Since the 1000 0000 0000 0000 0000 0000 0000 0000 is actually the Minvalue, this should also work:

<pre><code> int y = x | int.MinValue; // or, to do it to itself,
 x |= int.MinValue;
</code></pre>

blocks|key|1465962|text|只是为了更有趣：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465963|int+myInt+=+Math.Min(hisInt,+-hisInt);

int+myInt+=+-(int)Math.Sqrt(Math.Pow(Math.Sin(1),+2)+%2B+Math.Pow(Math.Cos(-1),+2))
++++++++++++*+Math.Abs(hisInt);|code-block|syntax|javascript|1465964|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|$]]

Just for more fun:

<pre><code>int myInt = Math.Min(hisInt, -hisInt);

int myInt = -(int)Math.Sqrt(Math.Pow(Math.Sin(1), 2) + Math.Pow(Math.Cos(-1), 2))
 * Math.Abs(hisInt);
</code></pre>

blocks|key|1251193|text|尽管我在这里的聚会来得太晚了，我还是要用我在硬件时代的一些有用的技巧加入进来。所有这些都假设有符号数的2的恭维表示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251194|int+negate+=+~i%2B1;
int+positiveMagnitude+=+(i+%5E+(i>>31))+-+(i>>31);
int+negativeMagnitude+=+(i>>31)+-+(i+%5E+(i>>31));|code-block|syntax|javascript|1251195|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|$]]

Even though I'm way late to the party here, I'm going to chime in with some useful tricks from my hardware days. All of these assume 2's compliment representation for signed numbers.

<pre><code>int negate = ~i+1;
int positiveMagnitude = (i ^ (i&gt;&gt;31)) - (i&gt;&gt;31);
int negativeMagnitude = (i&gt;&gt;31) - (i ^ (i&gt;&gt;31));
</code></pre>

blocks|key|1465918|text|也许是这个？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465919|int+n;

....+some+coding....

n+=+n<=0?+n:0-n;|code-block|syntax|javascript|1465920|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|$]]

Maybe this?

<pre><code>int n;

.... some coding....

n = n&lt;=0? n:0-n;
</code></pre>

blocks|key|1250900|text|int+myInt+=+-+System.Math.Abs(-5);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1250901|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>int myInt = - System.Math.Abs(-5);
</code></pre>

blocks|key|1465710|text|将其乘以-1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465711|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|$]]

blocks|key|1251683|text|正如前面提到的，仅仅乘以-1并不酷，因为int.MinValue+*+-1+==+int.MinValue|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1251684|这真的取决于你的应用，但对于我的(反转游戏杆轴)，我想确保所有的极端仍然是可以到达的，并保持尽可能多的保真度。|1251685|为此，我映射了最大<-->最小值，中间的所有内容都得到了一个*+-1|1251686|使用这种技术，在反转时-32767是不可达的。|1251687|++++private+static+int+Invert(int+value)
++++{
++++++++if+(value+==+int.MaxValue)+return+int.MinValue;
++++++++if+(value+==+int.MinValue)+return+int.MaxValue;
++++++++return+value+*+-1;
++++}|code-block|syntax|javascript|1251688|entityMap^0|K|X|0|0|U|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|Z|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|10|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

As previously mentioned, just multiplying by -1 is not cool, as <code>int.MinValue * -1 == int.MinValue</code>
It really depends on your application, but for mine (Inverting Joystick axes) I want to ensure that all extremes are still reachable, and maintain as much fidelity as possible. 
To this end, I map Max &lt;--&gt; Min and everything in-between gets a <code>* -1</code> 
Using this technique, -32767 is unreachable when inverting.
<pre><code> private static int Invert(int value)
 {
 if (value == int.MaxValue) return int.MinValue;
 if (value == int.MinValue) return int.MaxValue;
 return value * -1;
 }
</code></pre>

blocks|key|1251483|text|将数字从正数转换为负数，或将负数转换为正数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251484|public+static+decimal+Reverse(this+decimal+source)
{
+++++return+source+*+decimal.MinusOne;
}|code-block|syntax|javascript|1251485|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|$]]

Converting a number from positive to negative, or negative to positive:

<pre><code>public static decimal Reverse(this decimal source)
{
 return source * decimal.MinusOne;
}
</code></pre>

blocks|key|1466355|text|只是一些我想分享的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1466356|public+decimal+NumberReevaluate(decimal+number,+bool+isPositive)
{
++++var+tempNumber+=+System.Math.Abs(number);
++++return+isPositive+?+tempNumber+:+-tempNumber;
}|code-block|syntax|javascript|1466357|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|$]]

Just something that I want to share.

<pre><code>public decimal NumberReevaluate(decimal number, bool isPositive)
{
 var tempNumber = System.Math.Abs(number);
 return isPositive ? tempNumber : -tempNumber;
}
</code></pre>

blocks|key|1466166|text|使用二进制和删除负责负号的最后一位。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1466167|或者使用二进制或将符号添加到数据类型。|1466168|这个解决方案可能听起来荒唐和不完整，但我可以保证这是最快的方法。|1466169|如果你不尝试我发布的内容，这篇文章可能看起来很垃圾:D|1466170|例如int：|1466171|Int是32位数据类型，因此最后一位(第32位)决定符号。|1466172|32位的值为0，其余的为1。它会将负的no转换为%2Bve。|1466173|恰好相反，或者值为1，第32位，其余为0。|1466174|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|X|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|Y|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

Use binary and to remove the last bit which is responsible for negative sign.

Or use binary or to add sign to a datatype.

This soln may sound absurd and incomplete but I can guarantee this is the fastest method.

If you don't experiment with what I have posted this post may look crap :D

Eg for int:

Int is 32 bit datatype so the last bit (32th one) determines the sign.

And with a value which has 0 in the 32 place and rest 1. It will convert negative no to +ve.

For just the opposite or with a value with 1 in 32th place and rest 0.

blocks|key|1251269|text|X=*-1可能无法在所有编译器上运行...因为它读取的是‘乘法’‘减法’1而不是负数，所以更好的alt是X=(0-X)，它不同于X-=X|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1251270|entityMap^0|0|5|1G|7|1S|4|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

<code>X=*-1</code> may not work on all compilers... since it reads a 'multiply' 'SUBTRACT' 1 instead of NEGATIVE
The better alt is <code>X=(0-X)</code>, [WHICH IS DIFF FROM <code>X-=X</code>]

You can convert a negative number to positive like this:

<pre><code>int myInt = System.Math.Abs(-5);
</code></pre>

Is there an equivalent method to make a positive number negative?

Convert a positive number to negative in C#

可以将负数转换为正数，如下所示：int myInt = System.Math.Abs(-5);有没有等价的方法可以把一个正数变成负数？

问在C#中将正数转换为负数
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回答 19

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