blocks|key|1761866|text|Math.abs(num)+=>+Always+positive
-Math.abs(num)+=>+Always+negative|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1761867|但是，您确实意识到，对于您的代码，|unstyled|1761868|if($this.find('.pdxslide-activeSlide').index()+<+slideNum-1){+slideNum+=+-slideNum+}
console.log(slideNum)|1761869|如果找到的索引是3并且slideNum是3，|1761870|则3<+3-1个=>为假|1761871|所以slideNum+仍然是正的？？|offset|length|style|BOLD|1761872|在我看来，这更像是一个逻辑错误。|1761873|entityMap^0|0|0|0|0|0|B|5|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|11|8|@$O|12|P|13|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|F|7|15|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>Math.abs(num) =&gt; Always positive
-Math.abs(num) =&gt; Always negative
</code></pre>

You do realize however, that for your code

<pre><code>if($this.find('.pdxslide-activeSlide').index() &lt; slideNum-1){ slideNum = -slideNum }
console.log(slideNum)
</code></pre>

If the index found is 3 and slideNum is 3, 
then 3 &lt; 3-1 => false 
so slideNum remains positive??

It looks more like a logic error to me.

blocks|key|1761834|text|abs的反义词是Math.abs(num)+*+-1。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1761835|entityMap^0|8|I|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

The reverse of abs is <code>Math.abs(num) * -1</code>.

blocks|key|3834249|text|将正反转为负或将负反转为正的基本公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3834250|i+-+(i+*+2)|code-block|syntax|javascript|3834251|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 basic formula to reverse positive to negative or negative to positive:

<pre><code>i - (i * 2)
</code></pre>

blocks|key|144179|text|要在JavaScript中获得数字的负值，可以始终使用~位运算符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|144180|例如，如果您有a+=+1000，并且需要将其转换为负数，则可以执行以下操作：|144181|a+=+~a+%2B+1;|code-block|syntax|javascript|144182|这将导致a为-1000。|144183|entityMap^0|R|1|0|7|8|0|0|4|1|6|5|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|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

To get a negative version of a number in JavaScript you can always use the <code>~</code> bitwise operator.

For example, if you have <code>a = 1000</code> and you need to convert it to a negative, you could do the following:

<pre><code>a = ~a + 1;
</code></pre>

Which would result in <code>a</code> being <code>-1000</code>.

blocks|key|144280|text|var+x+=+100;
var+negX+=+(+-x+);+//+=>+-100|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|144281|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|$]]

blocks|key|1761817|text|您确定控制将进入if的主体吗？例如，if中的条件是否为真？因为如果不这样做，if的主体将永远不会被执行，并且slideNum将保持为正。我将冒险猜测这可能就是你所看到的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1761818|如果我在Firebug中尝试以下操作，它似乎可以工作：|1761819|>>>+i+=+5;+console.log(i);+i+=+-i;+console.log(i);
5
-5|code-block|syntax|javascript|1761820|slideNum+*=+-1也应该可以工作。Math.abs(slideNum)+*+-1也应该如此。|1761821|entityMap^0|8|2|I|2|12|2|1I|8|0|0|0|0|E|M|N|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

Are you sure that control is going into the body of the <code>if</code>? As in does the condition in the <code>if</code> ever hold true? Because if it doesn't, the body of the <code>if</code> will never get executed and <code>slideNum</code> will remain positive. I'm going to hazard a guess that this is probably what you're seeing.

If I try the following in Firebug, it seems to work:

<pre><code>&gt;&gt;&gt; i = 5; console.log(i); i = -i; console.log(i);
5
-5
</code></pre>

<code>slideNum *= -1</code> should also work. As should <code>Math.abs(slideNum) * -1</code>.

blocks|key|3834485|text|Javascript有一个专门的运算符：unary+negation。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3834486|戴利:这是减号！|3834487|要求反一个数字，只需以最直观的方式在其前面加上-即可。不需要编写函数，使用Math.abs()乘以-1或使用按位运算符。|style|CODE|3834488|一元否定适用于数字字面量：|3834489|let+a+=+10;++//+a+is+`10`
let+b+=+-10;+//+b+is+`-10`|code-block|syntax|javascript|3834490|它也适用于变量：|3834491|let+x+=+50;
x+=+-x;++++++//+x+is+now+`-50`

let+y+=+-6;
y+=+-y;++++++//+y+is+now+`6`|3834492|如果使用分组运算符，您甚至可以多次使用它。括号：|3834493|l+=+10;+++++++//+l+is+`10`
m+=+-10;++++++//+m+is+`-10`
n+=+-(10);++++//+n+is+`-10`
o+=+-(-(10));+//+o+is+`10`
p+=+-(-10);+++//+p+is+`10`+(double+negative+makes+a+positive)|3834494|以上所有操作也适用于变量。|3834495|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Unary_negation^0|K|E|0|0|0|N|1|11|A|1D|2|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|D|3|E|5|6|7|1C|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|1D|8|@$A|1E|B|1F|H|I]|$A|1G|B|1H|H|I]|$A|1I|B|1J|H|I]]|9|@]|C|$]]|$1|J|3|K|5|6|7|1K|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|1L|8|@]|9|@]|C|$O|P]]|$1|Q|3|R|5|6|7|1M|8|@]|9|@]|C|$]]|$1|S|3|T|5|N|7|1N|8|@]|9|@]|C|$O|P]]|$1|U|3|V|5|6|7|1O|8|@]|9|@]|C|$]]|$1|W|3|X|5|N|7|1P|8|@]|9|@]|C|$O|P]]|$1|Y|3|Z|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|10|3|-4|5|6|7|1R|8|@]|9|@]|C|$]]]|11|$12|$5|13|14|15|C|$16|17]]]]

Javascript has a dedicated operator for this: <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Unary_negation" rel="noreferrer">unary negation</a>. 

TL;DR: It's the minus sign! 

To negate a number, simply prefix it with <code>-</code> in the most intuitive possible way. No need to write a function, use <code>Math.abs()</code> multiply by <code>-1</code> or use the bitwise operator.

Unary negation works on number literals:

<pre><code>let a = 10; // a is `10`
let b = -10; // b is `-10`
</code></pre>

It works with variables too:

<pre><code>let x = 50;
x = -x; // x is now `-50`

let y = -6;
y = -y; // y is now `6`
</code></pre>

You can even use it multiple times if you use the grouping operator (a.k.a. parentheses:

<pre><code>l = 10; // l is `10`
m = -10; // m is `-10`
n = -(10); // n is `-10`
o = -(-(10)); // o is `10`
p = -(-10); // p is `10` (double negative makes a positive)
</code></pre>

All of the above works with a variable as well.

blocks|key|144240|text|如果不想使用Math.Abs+*+-1，可以使用这个简单的If语句:P|type|unstyled|depth|inlineStyleRanges|entityRanges|data|144241|if+(x+>+0)+{
++++x+=+-x;
}|code-block|syntax|javascript|144242|当然，您可以将此函数设置为如下所示|144243|function+makeNegative(number)+{
++++if+(number+>+0)+{
++++++++number+=+-number;
++++}
}|144244|makeNegative(+-3+)+=>+-3+makeNegative(5)+=>+-5|144245|希望这能有所帮助！Math.abs可能会为您工作，但如果它不是这样的话|144246|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you don't feel like using Math.Abs * -1 you can you this simple if statement :P

<pre><code>if (x &gt; 0) {
 x = -x;
}
</code></pre>

Of course you could make this a function like this

<pre><code>function makeNegative(number) {
 if (number &gt; 0) {
 number = -number;
 }
}
</code></pre>

makeNegative(-3) => -3
makeNegative(5) => -5

Hope this helps! Math.abs will likely work for you but if it doesn't this little

blocks|key|2236851|text|num+*+-1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2236852|这对你来说已经足够了。|unstyled|2236853|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>num * -1
</code></pre>

This would do it for you.

blocks|key|1762063|text|var+i+=+10;
i+=+i+/+-1;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1762064|结果：-10|unstyled|offset|length|style|CODE|1762065|var+i+=+-10;
i+=+i+/+-1;|1762066|结果：10|1762067|如果你除以负1，它总是会使你的数字反转。|1762068|entityMap^0|0|3|3|0|0|3|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@$G|U|H|V|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|X|8|@$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>var i = 10;
i = i / -1;
</code></pre>

Result: <code>-10</code>

<pre><code>var i = -10;
i = i / -1;
</code></pre>

Result: <code>10</code>

If you divide by negative 1, it will always flip your number either way.

blocks|key|3834348|text|使用0-x|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3834349|X是要反转的数字|3834350|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Use 0 - x

x being the number you want to invert

blocks|key|144402|text|它会将负数组转换为正数组，反之亦然|type|unstyled|depth|inlineStyleRanges|entityRanges|data|144403|function+negateOrPositive(arr)+{
++++arr.map(res+=>+-res)
};|code-block|syntax|javascript|144404|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|$]]

It will convert negative array to positive or vice versa

<pre><code>function negateOrPositive(arr) {
 arr.map(res =&gt; -res)
};
</code></pre>

blocks|key|1762090|text|在vanilla+javascript中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1762091|​|1762092|if(number+>+0)
++return+-1*number;|code-block|syntax|javascript|1762093|1762094|其中，上面的是您想要转换的正数|1762095|此代码将通过乘以-1简单地将正数转换为负数|1762096|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|T|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|U|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|O|$]]

In vanilla javascript 

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>if(number &gt; 0)
 return -1*number;</code></pre>
</div>
</div>


Where number above is the positive number you intend to convert 

This code will convert just positive numbers to negative numbers simple by multiplying by -1

Basically, the reverse of abs. If I have:

<pre><code>if ($this.find('.pdxslide-activeSlide').index() &lt; slideNum - 1) {
 slideNum = -slideNum
}
console.log(slideNum)
</code></pre>

No matter what console always returns a positive number. How do I fix this?

If I do:

<pre><code>if ($this.find('.pdxslide-activeSlide').index() &lt; slideNum - 1) {
 _selector.animate({
 left: (-slideNum * sizes.images.width) + 'px'
 }, 750, 'InOutPDX')
} else {
 _selector.animate({
 left: (slideNum * sizes.images.width) + 'px'
 }, 750, 'InOutPDX')
}
</code></pre>

it works tho, but it's not "DRY" and just stupid to have an entire block of code JUST for a <code>-</code>.

Positive Number to Negative Number in JavaScript?

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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基本上就是腹肌的反面。如果我有：if ($this.find('.pdxslide-activeSlide').index() < slideNum - 1) { slideNum = -slideNum}console.log(slideNum)无论哪种类型的控制台都会返回一个正数。我该如何解决这个问题？如果我这样做了：if ($this.find('.pdxslide-activeSlide

问在JavaScript中将正数转换为负数？
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回答 13

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问在JavaScript中将正数转换为负数？
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