blocks|key|1372334|text|尝试BigDecimal的setScale，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1372335|public+static+double+round(double+d,+int+decimalPlace)+{
++++BigDecimal+bd+=+new+BigDecimal(d);
++++bd+=+bd.setScale(decimalPlace,+BigDecimal.ROUND_HALF_UP);
++++return+bd.doubleValue();
}|code-block|syntax|javascript|1372336|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html^0|2|A|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Try setScale of <a href="http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html" rel="noreferrer">BigDecimal</a> like so:

<pre><code>public static double round(double d, int decimalPlace) {
 BigDecimal bd = new BigDecimal(d);
 bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
 return bd.doubleValue();
}
</code></pre>

blocks|key|1157445|text|令人难以置信的是，现在还没有人提出这一点，Java+API已经有很多年的DecimalFormat就是为了这个目的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1157446|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.4.2/docs/api/java/text/DecimalFormat.html^0|10|D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Incredible no one brought this up yet, Java API has had <a href="http://java.sun.com/j2se/1.4.2/docs/api/java/text/DecimalFormat.html" rel="noreferrer">DecimalFormat</a> for ages now for this exact purpose.

blocks|key|1157161|text|对于大多数数字，除非使用有保证精度的十进制类，如BigDecimal，否则无法获得xxx.yyyy的精确表示。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1157162|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html^0|15|8|O|A|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

For most numbers, you won't be able to get an exact representation of <code>xxx.yyyy</code> unless you use a decimal class with guaranteed accuracy, such as <a href="http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html" rel="noreferrer">BigDecimal</a>.

blocks|key|1157577|text|只需删除小数部分|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1157578|+`public+double+trunk(double+value){+++++++++return+value+-+value+%25+1;+++++}`|code-block|syntax|javascript|1157579|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|$]]

Simply remove the fractional portion

<code>
 public double trunk(double value){
 return value - value % 1;
 }
</code>

blocks|key|1372578|text|下面是一个简短的实现，它比使用BigDecimal或Math.pow快许多倍|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1372579|private+static+long+TENS[]+=+new+long[19];
static+{
++++TENS[0]+=+1;
++++for+(int+i+=+1;+i+<+TENS.length;+i%2B%2B)+TENS[i]+=+10+*+TENS[i+-+1];
}

public+static+double+round(double+v,+int+precision)+{
++++assert+precision+>=+0+&&+precision+<+TENS.length;
++++double+unscaled+=+v+*+TENS[precision];
++++if(unscaled+<+Long.MIN_VALUE+%7C%7C+unscaled+>+Long.MAX_VALUE)+
+++++++return+v;
++++long+unscaledLong+=+(long)+(unscaled+%2B+(v+<+0+?+-0.5+:+0.5));
++++return+(double)+unscaledLong+/+TENS[precision];
}|code-block|syntax|javascript|1372580|删除要品尝的断言。;)|1372581|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|$]]

here is a short implementation which is many times faster than using BigDecimal or Math.pow

<pre><code>private static long TENS[] = new long[19];
static {
 TENS[0] = 1;
 for (int i = 1; i &lt; TENS.length; i++) TENS[i] = 10 * TENS[i - 1];
}

public static double round(double v, int precision) {
 assert precision &gt;= 0 &amp;&amp; precision &lt; TENS.length;
 double unscaled = v * TENS[precision];
 if(unscaled &lt; Long.MIN_VALUE || unscaled &gt; Long.MAX_VALUE) 
 return v;
 long unscaledLong = (long) (unscaled + (v &lt; 0 ? -0.5 : 0.5));
 return (double) unscaledLong / TENS[precision];
}
</code></pre>

Delete the assert'ions to taste. ;)

blocks|key|1372600|text|使用这个简单的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1372601|double+truncateDouble(double+number,+int+numDigits)+{
++++double+result+=+number;
++++String+arg+=+""+%2B+number;
++++int+idx+=+arg.indexOf('.');
++++if+(idx!=-1)+{
++++++++if+(arg.length()+>+idx%2BnumDigits)+{
++++++++++++arg+=+arg.substring(0,idx%2BnumDigits%2B1);
++++++++++++result++=+Double.parseDouble(arg);
++++++++}
++++}
++++return+result+;
}|code-block|syntax|javascript|1372602|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|$]]

Use this simple function

<pre><code>double truncateDouble(double number, int numDigits) {
 double result = number;
 String arg = "" + number;
 int idx = arg.indexOf('.');
 if (idx!=-1) {
 if (arg.length() &gt; idx+numDigits) {
 arg = arg.substring(0,idx+numDigits+1);
 result = Double.parseDouble(arg);
 }
 }
 return result ;
}
</code></pre>

blocks|key|1372637|text|我只想补充ubuntudroid的解决方案。我试过了，它不会四舍五入，所以我不得不添加|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1372638|df.setRoundingMode(RoundingMode.FLOOR);|code-block|syntax|javascript|1372639|才能正常工作。|1372640|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|$]]

I just want to add to ubuntudroid's solution.
I tried it and it wouldn't round down, so I had to add

<pre><code>df.setRoundingMode(RoundingMode.FLOOR);
</code></pre>

for it to work.

blocks|key|1157232|text|要做到100%25可靠，你必须将参数作为字符串传递，而不是作为浮点数。当以字符串形式给出时，代码很容易编写。这样的原因是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1157233|double+x+=+1.1;|code-block|syntax|javascript|1157234|并不意味着x实际上将计算为1.1，而只是最接近可精确表示的数字。|1157235|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|$]]

To do it 100% reliably, you'd have to pass the argument as string, not as floating-point number. When given as string, the code is easy to write. The reason for this is that 

<pre><code>double x = 1.1;
</code></pre>

does not mean that x will actually evaluate to 1.1, only to the closest exactly representable number.

blocks|key|1372459|text|实际上，这类东西很容易编写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1372460|public+static+double+truncate(double+value,+int+places)+{
++++double+multiplier+=+Math.pow(10,+places);
++++return+Math.floor(multiplier+*+value)+/+multiplier;
}|code-block|syntax|javascript|1372461|注意，它是Math.floor，因为Math.round不会截断。|1372462|哦，这会返回一个双精度值，因为数学类中的大多数函数(如Math.pow和Math.floor)都会返回双精度值。|1372463|警告:替身在准确性方面很糟糕。首先应该考虑BigDecimal解决方案之一。|1372464|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Actually, this sort of thing is easy to write:

<pre><code>public static double truncate(double value, int places) {
 double multiplier = Math.pow(10, places);
 return Math.floor(multiplier * value) / multiplier;
}
</code></pre>

Note that it's Math.floor, because Math.round wouldn't be truncating.

Oh, and this returns a double, because that's what most functions in the Math class return (like Math.pow and Math.floor).

Caveat: Doubles suck for accuracy. One of the BigDecimal solutions should be considered first.

blocks|key|1157333|text|创建了一个方法来做这件事。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1157334|public+double+roundDouble(double+d,+int+places)+{
++++return+Math.round(d+*+Math.pow(10,+(double)+places))+/+Math.pow(10,+(double)places);
}|code-block|syntax|javascript|1157335|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|$]]

created a method to do it.

<pre><code>public double roundDouble(double d, int places) {
 return Math.round(d * Math.pow(10, (double) places)) / Math.pow(10, (double)places);
}
</code></pre>

Is there a Java Library function which can be used to truncate a number to an arbitrary number of decimal places?
For Example.

<pre><code>SomeLibrary.truncate(1.575, 2) = 1.57
</code></pre>

Thanks

Are there any functions for truncating a double in java?

Java

有没有一个Java库函数可以用来将一个数字截断到任意数量的小数位？例如。SomeLibrary.truncate(1.575, 2) = 1.57谢谢

问java中有没有截断双精度值的函数？
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问java中有没有截断双精度值的函数？
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