blocks|key|1423566|text|打电话给Math.abs就行了。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1423567|int+x+=+Math.abs(-5);|code-block|syntax|javascript|1423568|这会将x设置为5。|style|CODE|1423569|请注意，如果您传递Integer.MIN_VALUE，将返回相同的值(仍然为负值)，因为int的范围不允许表示正等效值。|1423570|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#abs-int-^0|4|8|0|0|0|3|1|7|1|0|9|H|18|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@$A|12|B|13|K|L]|$A|14|B|15|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|16|8|@$A|17|B|18|K|L]|$A|19|B|1A|K|L]]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1B|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

Just call <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#abs-int-" rel="noreferrer">Math.abs</a>. For example:
<pre><code>int x = Math.abs(-5);
</code></pre>
Which will set <code>x</code> to <code>5</code>.
Note that if you pass <code>Integer.MIN_VALUE</code>, the same value (still negative) will be returned, as the range of <code>int</code> does not allow the positive equivalent to be represented.

blocks|key|1423673|text|您正在描述的概念称为“绝对值”，Java有一个名为Math.abs的函数来为您做这件事。或者你可以避免函数调用，自己去做：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1423674|number+=+(number+<+0+?+-number+:+number);|code-block|syntax|javascript|1423675|或|1423676|if+(number+<+0)
++++number+=+-number;|1423677|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/javase/6/docs/api/java/lang/Math.html#abs(int)^0|P|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

The concept you are describing is called "absolute value", and Java has a function called <a href="http://java.sun.com/javase/6/docs/api/java/lang/Math.html#abs(int)" rel="noreferrer">Math.abs</a> to do it for you. Or you could avoid the function call and do it yourself:

<pre><code>number = (number &lt; 0 ? -number : number);
</code></pre>

or 

<pre><code>if (number &lt; 0)
 number = -number;
</code></pre>

blocks|key|1638706|text|你要的是绝对值，伙计。Math.abs(-5)返回5...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1638707|entityMap^0|B|C|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|$]]

You're looking for absolute value, mate. <code>Math.abs(-5)</code> returns 5...

blocks|key|1638633|text|使用abs函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1638634|int+sum=0;
for(Integer+i+:+container)
++sum%2B=Math.abs(i);|code-block|syntax|javascript|1638635|entityMap^0|2|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Use the <code>abs</code> function:

<pre><code>int sum=0;
for(Integer i : container)
 sum+=Math.abs(i);
</code></pre>

blocks|key|1423610|text|你是在问绝对值吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1423611|Math.abs(...)是您可能想要的函数。|1423612|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|$]]

Are you asking about absolute values?

Math.abs(...) is the function you probably want.

blocks|key|1423814|text|您希望将每个数字包装到Math.abs()中。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1423815|System.out.println(Math.abs(-1));|code-block|syntax|javascript|1423816|打印"1“。|1423817|如果您希望避免编写数学工具，可以静态地包含Math.-part。只需写|1423818|import+static+java.lang.Math.abs;|1423819|和导入一起使用，您只需编写以下代码即可引用abs()-function|1423820|System.out.println(abs(-1));|1423821|entityMap^0|B|A|0|0|0|L|5|0|0|L|5|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|V|$]]

You want to wrap each number into <code>Math.abs()</code>. e.g.

<pre><code>System.out.println(Math.abs(-1));
</code></pre>

prints out "1".

If you want to avoid writing the <code>Math.</code>-part, you can include the Math util statically. Just write 

<pre><code>import static java.lang.Math.abs;
</code></pre>

along with your imports, and you can refer to the <code>abs()</code>-function just by writing

<pre><code>System.out.println(abs(-1));
</code></pre>

blocks|key|1638676|text|最简单的方法是将每个数字包装在一个Math.abs()调用中，因此您可以添加：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1638677|Math.abs(1)+%2B+Math.abs(2)+%2B+Math.abs(1)+%2B+Math.abs(-1)|code-block|syntax|javascript|1638678|通过逻辑更改来反映您的代码是如何构造的。也许是冗长的，但它做了你想要的。|1638679|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|$]]

The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:

<pre><code>Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)
</code></pre>

with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.

blocks|key|1423964|text|如果你对2的补码机制感兴趣，这里是绝对低效的，但说明性的低级方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1423965|private+static+int+makeAbsolute(int+number){
+++++if(number+>=0){
++++++++return+number;
+++++}+else{
++++++++return+(~number)%2B1;
+++++}
}|code-block|syntax|javascript|1423966|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|$]]

If you're interested in the mechanics of two's complement, here's the absolutely inefficient, but illustrative low-level way this is made:

<pre><code>private static int makeAbsolute(int number){
 if(number &gt;=0){
 return number;
 } else{
 return (~number)+1;
 }
}
</code></pre>

blocks|key|1423897|text|String+s+=+"-1139627840";
BigInteger+bg1+=+new+BigInteger(s);
System.out.println(bg1.abs());|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1423898|或者：|unstyled|1423899|int+i+=+-123;
System.out.println(Math.abs(i));|1423900|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>String s = "-1139627840";
BigInteger bg1 = new BigInteger(s);
System.out.println(bg1.abs());
</code></pre>

Alternatively:

<pre><code>int i = -123;
System.out.println(Math.abs(i));
</code></pre>

blocks|key|1423894|text|我需要long的绝对值，并深入研究了Math.abs，发现如果我的参数小于-9223372036854775808l的LONG.MIN_VAL，那么abs函数将不会返回绝对值，而只会返回最小值。在这种情况下，如果您的代码进一步使用这个abs值，那么可能会出现问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1423895|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|$]]

I needed the absolute value of a long , and looked deeply into Math.abs and found that if my argument is less than LONG.MIN_VAL which is -9223372036854775808l, then the abs function would not return an absolute value but only the minimum value. Inthis case if your code is using this abs value further then there might be an issue.

blocks|key|1424035|text|你能试试这个吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1424036|public+static+int+toPositive(int+number)+{
++++return+number+&+0x7fffffff;
}|code-block|syntax|javascript|1424037|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|$]]

Can you please try this one? 

<pre><code>public static int toPositive(int number) {
 return number &amp; 0x7fffffff;
}
</code></pre>

blocks|key|1638974|text|在kotlin中，您可以使用unaryPlus|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1638975|input+=+input.unaryPlus()|code-block|syntax|javascript|1638976|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html|offset|length|1638977|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|20|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

In kotlin you can use unaryPlus

<pre><code>input = input.unaryPlus()
</code></pre>

<a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html" rel="nofollow noreferrer">https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html</a>

blocks|key|1638995|text|if(arr[i]<0)

Math.abs(arr[i]);++//1st+way+(taking+absolute+value)

arr[i]=-(arr[i]);+//2nd+way+(taking+-ve+of+-ve+no.+yields+a+%2Bve+no.)

arr[i]=+~(arr[i]-1);+++//3rd+way+(taking+negation)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1638996|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>if(arr[i]&lt;0)

Math.abs(arr[i]); //1st way (taking absolute value)

arr[i]=-(arr[i]); //2nd way (taking -ve of -ve no. yields a +ve no.)

arr[i]= ~(arr[i]-1); //3rd way (taking negation)
</code></pre>

blocks|key|1424103|text|我看到人们在说Math.abs(number)，但这种方法并不是完全的证据。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1424104|当您尝试包装Math.abs(Integer.MIN_VALUE)时，此操作失败(请参阅https://youtu.be/IWrpDP-ad7g)|1424105|如果您不确定是否会收到输入中的Integer.MIN_VALUE。始终建议检查该号码并手动处理它。|1424106|entityMap|0|LINK|mutability|MUTABLE|url|https://youtu.be/IWrpDP-ad7g^0|7|G|0|6|R|18|S|0|0|F|H|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|I|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

I see people are saying that <code>Math.abs(number)</code> but this method is not full proof.
This fails when you try to wrap <code>Math.abs(Integer.MIN_VALUE)</code> (see ref. <a href="https://youtu.be/IWrpDP-ad7g" rel="nofollow noreferrer">https://youtu.be/IWrpDP-ad7g</a>)
If you are not sure whether you are going to receive the <code>Integer.MIN_VALUE</code> in the input. It is always recommended to check for that number and handle it manually.

blocks|key|1424146|text|在for循环中尝试这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1424147|sum+%2B=+Math.abs(arr[i])|code-block|syntax|javascript|1424148|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|$]]

Try this in the for loop:
<pre><code>sum += Math.abs(arr[i])
</code></pre>

blocks|key|1638918|text|不要这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1638919|number+=+(number+<0-+-number+:+number)；|1638920|或|1638921|如果(number+<+0)+number+=+-number；|1638922|这将是一个错误，当你在你的代码上运行find+bug时，它会报告为RV_NEGATING_RESULT_OF|1638923|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

dont do this

number = (number &lt; 0 ? -number : number);

or

if (number &lt; 0) number = -number;

this will be an bug when you run find bug on your code it will report it as RV_NEGATING_RESULT_OF

I have a Java method in which I'm summing a set of numbers. However, I want any negatives numbers to be treated as positives. So (1)+(2)+(1)+(-1) should equal 5. 

I'm sure there is very easy way of doing this - I just don't know how.

Make a negative number positive

Java

我有一个对一组数字求和的Java方法。但是，我希望任何负数都被视为正数。所以(1)+(2)+(1)+(-1)应该等于5。我相信有一种非常简单的方法可以做到这一点--我只是不知道怎么做。

问使负数变为正数
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回答 16

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