blocks|key|1465493|text|假设您对截断到零很满意，那么就强制转换：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465494|double+d+=+1234.56;
long+x+=+(long)+d;+//+x+=+1234|code-block|syntax|javascript|1465495|这将比通过包装器类更快-更重要的是，它更具可读性。现在，如果您需要舍入而不是“始终接近零”，那么您将需要稍微复杂一些的代码。|1465496|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|$]]

Assuming you're happy with truncating towards zero, just cast:

<pre><code>double d = 1234.56;
long x = (long) d; // x = 1234
</code></pre>

This will be faster than going via the wrapper classes - and more importantly, it's more readable. Now, if you need rounding other than "always towards zero" you'll need slightly more complicated code.

blocks|key|1465533|text|..。这是不截断的舍入方法。急忙在Java+API手册中查找它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465534|double+d+=+1234.56;
long+x+=+Math.round(d);+//1235|code-block|syntax|javascript|1465535|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|$]]

... And here is the rounding way which doesn't truncate. Hurried to look it up in the Java API Manual:

<pre><code>double d = 1234.56;
long x = Math.round(d); //1235
</code></pre>

blocks|key|1680856|text|(new+Double(d)).longValue()只在内部执行强制转换，所以没有理由创建一个双对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1680857|entityMap^0|0|R|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|$]]

<code>(new Double(d)).longValue()</code> internally just does a cast, so there's no reason to create a Double object.

blocks|key|1465681|text|Guava+Math库有一个专门为将双精度转换为长整型而设计的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465682|long+DoubleMath.roundToLong(double+x,+RoundingMode+mode)|code-block|syntax|javascript|1465683|可以使用java.math.RoundingMode指定舍入行为。|offset|length|style|CODE|1465684|entityMap^0|0|0|4|M|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Guava Math library has a method specially designed for converting a double to a long:

<pre><code>long DoubleMath.roundToLong(double x, RoundingMode mode)
</code></pre>

You can use <code>java.math.RoundingMode</code> to specify the rounding behavior.

blocks|key|1465778|text|如果您强烈怀疑DOUBLE实际上是一个长整型，并且您想要|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465779|1)获取其确切值的句柄，作为LONG|1465780|2)当错误不长时抛出错误|1465781|您可以尝试如下所示：|1465782|public+class+NumberUtils+{

++++/**
++++*+Convert+a+{@link+Double}+to+a+{@link+Long}.
++++*+Method+is+for+{@link+Double}s+that+are+actually+{@link+Long}s+and+we+just
++++*+want+to+get+a+handle+on+it+as+one.
++++*/
++++public+static+long+getDoubleAsLong(double+specifiedNumber)+{
++++++++Assert.isTrue(NumberUtils.isWhole(specifiedNumber));
++++++++Assert.isTrue(specifiedNumber+<=+Long.MAX_VALUE+&&+specifiedNumber+>=+Long.MIN_VALUE);
++++++++//+we+already+know+its+whole+and+in+the+Long+range
++++++++return+Double.valueOf(specifiedNumber).longValue();
++++}

++++public+static+boolean+isWhole(double+specifiedNumber)+{
++++++++//+http://stackoverflow.com/questions/15963895/how-to-check-if-a-double-value-has-no-decimal-part
++++++++return+(specifiedNumber+%25+1+==+0);
++++}
}|code-block|syntax|javascript|1465783|Long是Double的一个子集，因此如果您无意中尝试转换超出Long范围的Double，您可能会得到一些奇怪的结果：|1465784|@Test
public+void+test()+throws+Exception+{
++++//+Confirm+that+LONG+is+a+subset+of+DOUBLE,+so+numbers+outside+of+the+range+can+be+problematic
++++Assert.isTrue(Long.MAX_VALUE+<+Double.MAX_VALUE);
++++Assert.isTrue(Long.MIN_VALUE+>+-Double.MAX_VALUE);+//+Not+Double.MIN_VALUE+=>+read+the+Javadocs,+Double.MIN_VALUE+is+the+smallest+POSITIVE+double,+not+the+bottom+of+the+range+of+values+that+Double+can+possible+be

++++//+Double.longValue()+failure+due+to+being+out+of+range+=>+results+are+the+same+even+though+I+minus+ten
++++System.out.println("Double.valueOf(Double.MAX_VALUE).longValue():+"+%2B+Double.valueOf(Double.MAX_VALUE).longValue());
++++System.out.println("Double.valueOf(Double.MAX_VALUE+-+10).longValue():+"+%2B+Double.valueOf(Double.MAX_VALUE+-+10).longValue());

++++//+casting+failure+due+to+being+out+of+range+=>+results+are+the+same+even+though+I+minus+ten
++++System.out.println("(long)+Double.valueOf(Double.MAX_VALUE):+"+%2B+(long)+Double.valueOf(Double.MAX_VALUE).doubleValue());
++++System.out.println("(long)+Double.valueOf(Double.MAX_VALUE+-+10).longValue():+"+%2B+(long)+Double.valueOf(Double.MAX_VALUE+-+10).doubleValue());
}|1465785|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

If you have a strong suspicion that the DOUBLE is actually a LONG, and you want to

1) get a handle on its EXACT value as a LONG

2) throw an error when its not a LONG

you can try something like this:

<pre><code>public class NumberUtils {

 /**
 * Convert a {@link Double} to a {@link Long}.
 * Method is for {@link Double}s that are actually {@link Long}s and we just
 * want to get a handle on it as one.
 */
 public static long getDoubleAsLong(double specifiedNumber) {
 Assert.isTrue(NumberUtils.isWhole(specifiedNumber));
 Assert.isTrue(specifiedNumber &lt;= Long.MAX_VALUE &amp;&amp; specifiedNumber &gt;= Long.MIN_VALUE);
 // we already know its whole and in the Long range
 return Double.valueOf(specifiedNumber).longValue();
 }

 public static boolean isWhole(double specifiedNumber) {
 // http://stackoverflow.com/questions/15963895/how-to-check-if-a-double-value-has-no-decimal-part
 return (specifiedNumber % 1 == 0);
 }
}
</code></pre>

Long is a subset of Double, so you might get some strange results if you unknowingly try to convert a Double that is outside of Long's range:

<pre><code>@Test
public void test() throws Exception {
 // Confirm that LONG is a subset of DOUBLE, so numbers outside of the range can be problematic
 Assert.isTrue(Long.MAX_VALUE &lt; Double.MAX_VALUE);
 Assert.isTrue(Long.MIN_VALUE &gt; -Double.MAX_VALUE); // Not Double.MIN_VALUE =&gt; read the Javadocs, Double.MIN_VALUE is the smallest POSITIVE double, not the bottom of the range of values that Double can possible be

 // Double.longValue() failure due to being out of range =&gt; results are the same even though I minus ten
 System.out.println("Double.valueOf(Double.MAX_VALUE).longValue(): " + Double.valueOf(Double.MAX_VALUE).longValue());
 System.out.println("Double.valueOf(Double.MAX_VALUE - 10).longValue(): " + Double.valueOf(Double.MAX_VALUE - 10).longValue());

 // casting failure due to being out of range =&gt; results are the same even though I minus ten
 System.out.println("(long) Double.valueOf(Double.MAX_VALUE): " + (long) Double.valueOf(Double.MAX_VALUE).doubleValue());
 System.out.println("(long) Double.valueOf(Double.MAX_VALUE - 10).longValue(): " + (long) Double.valueOf(Double.MAX_VALUE - 10).doubleValue());
}
</code></pre>

blocks|key|1465817|text|简单地通过以下几点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465818|double+d+=+394.000;
long+l+=+d+*+1L;|code-block|syntax|javascript|1465819|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 by the following:

<pre><code>double d = 394.000;
long l = d * 1L;
</code></pre>

blocks|key|1465635|text|简单地说，强制转换比创建一个双对象更有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465636|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|$]]

Simply put, casting is more efficient than creating a Double object.

What is the best way to convert a double to a long without casting?

For example:

<pre><code>double d = 394.000;
long l = (new Double(d)).longValue();
System.out.println("double=" + d + ", long=" + l);
</code></pre>

How to convert a double to long without casting?

Java

在不进行强制转换的情况下，将双精度型转换为长型的最佳方法是什么？例如：double d = 394.000;long l = (new Double(d)).longValue();System.out.println("double=" + d + ", long=" + l);

问如何在不进行强制转换的情况下将双精度型转换为长整型？
EN

回答 7

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问如何在不进行强制转换的情况下将双精度型转换为长整型？EN

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问如何在不进行强制转换的情况下将双精度型转换为长整型？
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