blocks|key|140922|text|是的，除以它。1+/+%2B0.0f是%2BInfinity，但1+/+-0.0f是-Infinity。通过简单的比较可以很容易地找出是哪一个，所以你会得到：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|140923|if+(1+/+x+>+0)
++++//+%2B0+here
else
++++//+-0+here|code-block|syntax|javascript|140924|(这假设x只能是两个零中的一个)|140925|entityMap^0|7|9|H|9|S|9|12|9|0|0|4|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

Yes, divide by it. <code>1 / +0.0f</code> is <code>+Infinity</code>, but <code>1 / -0.0f</code> is <code>-Infinity</code>. It's easy to find out which one it is with a simple comparison, so you get:

<pre><code>if (1 / x &gt; 0)
 // +0 here
else
 // -0 here
</code></pre>

(this assumes that <code>x</code> can only be one of the two zeroes)

blocks|key|140982|text|您可以使用Float.floatToIntBits将其转换为int并查看位模式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|140983|float+f+=+-0.0f;

if+(Float.floatToIntBits(f)+==+0x80000000)+{
++++System.out.println("Negative+zero");
}|code-block|syntax|javascript|140984|entityMap^0|5|K|U|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>Float.floatToIntBits</code> to convert it to an <code>int</code> and look at the bit pattern:

<pre><code>float f = -0.0f;

if (Float.floatToIntBits(f) == 0x80000000) {
 System.out.println("Negative zero");
}
</code></pre>

blocks|key|4647578|text|绝对不是最好的方法。签出函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4647579|Float.floatToRawIntBits(f);|code-block|syntax|javascript|4647580|Doku：|4647581|/**
+*+Returns+a+representation+of+the+specified+floating-point+value
+*+according+to+the+IEEE+754+floating-point+"single+format"+bit
+*+layout,+preserving+Not-a-Number+(NaN)+values.
+*
+*+Bit+31+(the+bit+that+is+selected+by+the+mask
+*+{@code+0x80000000})+represents+the+sign+of+the+floating-point
+*+number.
+...
+public+static+native+int+floatToRawIntBits(float+value);|4647582|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|$]]

Definitly not the best aproach. Checkout the function

<pre><code>Float.floatToRawIntBits(f);
</code></pre>

Doku:

<pre><code>/**
 * Returns a representation of the specified floating-point value
 * according to the IEEE 754 floating-point "single format" bit
 * layout, preserving Not-a-Number (NaN) values.
 *
 * &lt;p&gt;Bit 31 (the bit that is selected by the mask
 * {@code 0x80000000}) represents the sign of the floating-point
 * number.
 ...
 public static native int floatToRawIntBits(float value);
</code></pre>

blocks|key|141066|text|Math.min使用的方法类似于Jesper提出的方法，但更清晰一些：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|141067|private+static+int+negativeZeroFloatBits+=+Float.floatToRawIntBits(-0.0f);

float+f+=+-0.0f;
boolean+isNegativeZero+=+(Float.floatToRawIntBits(f)+==+negativeZeroFloatBits);|code-block|syntax|javascript|141068|entityMap^0|0|8|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|$]]

The approach used by <code>Math.min</code> is similar to what Jesper proposes but a little clearer:

<pre><code>private static int negativeZeroFloatBits = Float.floatToRawIntBits(-0.0f);

float f = -0.0f;
boolean isNegativeZero = (Float.floatToRawIntBits(f) == negativeZeroFloatBits);
</code></pre>

blocks|key|4647653|text|当浮点数为负(包括-0.0和-inf)时，它使用与负整数相同的符号位。这意味着您可以将整数表示与0进行比较，从而无需知道或计算-0.0的整数表示|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4647654|if(f+==+0.0)+{
++if(Float.floatToIntBits(f)+<+0)+{
++++//negative+zero
++}+else+{
++++//positive+zero
++}
}|code-block|syntax|javascript|4647655|这比公认的答案有一个额外的分支，但我认为没有十六进制常量，它更具可读性。|4647656|如果您的目标只是将-0视为负数，则可以省略外部if语句：|4647657|if(Float.floatToIntBits(f)+<+0)+{
++//any+negative+float,+including+-0.0+and+-inf
}+else+{
++//any+non-negative+float,+including+%2B0.0,+%2Binf,+and+NaN
}|4647658|entityMap^0|9|4|E|4|1C|1|1R|4|0|0|0|N|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

When a float is negative (including <code>-0.0</code> and <code>-inf</code>), it uses the same sign bit as a negative int. This means you can compare the integer representation to <code>0</code>, eliminating the need to know or compute the integer representation of <code>-0.0</code>:

<pre><code>if(f == 0.0) {
 if(Float.floatToIntBits(f) &lt; 0) {
 //negative zero
 } else {
 //positive zero
 }
}
</code></pre>

That has an extra branch over the accepted answer, but I think it's more readable without a hex constant.

If your goal is just to treat -0 as a negative number, you could leave out the outer <code>if</code> statement:

<pre><code>if(Float.floatToIntBits(f) &lt; 0) {
 //any negative float, including -0.0 and -inf
} else {
 //any non-negative float, including +0.0, +inf, and NaN
}
</code></pre>

blocks|key|141133|text|在Java中，Double.equals区分±0.0。(还有Float.equals。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|141134|我有点惊讶没有人提到这些，因为在我看来，它们比到目前为止给出的任何方法都要清晰！|141135|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/lang/Double.html#equals-java.lang.Object-|1|http://docs.oracle.com/javase/8/docs/api/java/lang/Float.html#equals-java.lang.Object-^0|7|D|U|C|7|D|0|U|C|1|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]|$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

<a href="http://docs.oracle.com/javase/8/docs/api/java/lang/Double.html#equals-java.lang.Object-"><code>Double.equals</code></a> distinguishes ±0.0 in Java. (There's also <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/Float.html#equals-java.lang.Object-"><code>Float.equals</code></a>.)

I'm a bit surprised no-one has mentioned these, as they seem to me clearer than any method given so far!

blocks|key|4647712|text|对于否定的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4647713|new+Double(-0.0).equals(new+Double(value));|code-block|syntax|javascript|4647714|积极的方面：|4647715|new+Double(0.0).equals(new+Double(value));|4647716|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|$]]

For negative: 

<pre><code>new Double(-0.0).equals(new Double(value));
</code></pre>

For positive: 

<pre><code>new Double(0.0).equals(new Double(value));
</code></pre>

blocks|key|4647753|text|只需使用Double.compare+(d1，d2)即可。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4647754|double+d1+=+-0.0;++++//+or+d1+=+0.0

if+(+Double.compare+(d1,+0.0)++<+0+)
++++System.out.println("negative+zero");
else
++++System.out.println("positive+zero");|code-block|syntax|javascript|4647755|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 use Double.compare (d1,d2).
<pre><code>double d1 = -0.0; // or d1 = 0.0

if ( Double.compare (d1, 0.0) &lt; 0 )
 System.out.println(&quot;negative zero&quot;);
else
 System.out.println(&quot;positive zero&quot;);
</code></pre>

Is it possible to check if a <code>float</code> is a positive zero (0.0) or a negative zero (-0.0)?

I've converted the <code>float</code> to a <code>String</code> and checked if the first <code>char</code> is a <code>'-'</code>, but are there any other ways?

How do I check if a zero is positive or negative?

Java

是否可以检查float是正零(0.0)还是负零(-0.0)？我已经将float转换为String，并检查了第一个char是否是'-'，但还有其他方法吗？

问如何检查零是正的还是负的？
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