How to convert a double to long without casting?

问题:

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

For example:

double d = 394.000; long l = (new Double(d)).longValue(); System.out.println("double=" + d + ", long=" + l); 

回答1:

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

double d = 1234.56; long x = (long) d; // x = 1234 

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.

回答2:

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

double d = 1234.56; long x = Math.round(d); 

回答3:

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

回答4:

The preferred approach should be:

Double.valueOf(d).longValue() 

From the Double (Java Platform SE 7) documentation:

Double.valueOf(d) 

Returns a Double instance representing the specified double value. If a new Double instance is not required, this method should generally be used in preference to the constructor Double(double), as this method is likely to yield significantly better space and time performance by caching frequently requested values.

回答5:

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

long DoubleMath.roundToLong(double x, RoundingMode mode) 

You can use java.math.RoundingMode to specify the rounding behavior.

回答6:

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:

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 % 1 == 0);     } } 

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:

@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(): " + 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 => 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()); } 

回答7:

Do you want to have a binary conversion like

double result = Double.longBitsToDouble(394.000d); 

回答8:

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

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