首选的方法应该是:

Double.valueOf(d).longValue()

来自Double (Java Platform SE 7)文档:

Double.valueOf(d)

返回表示指定Double值的Double实例。 如果不需要新的Double实例，则应该使用此方法 通常优先于构造函数Double(Double)， 因为这种方法很可能产生明显更好的空间和时间 缓存频繁请求值的性能。

假设你对截断为0感到满意，只需强制转换:

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

这将比使用包装器类更快——更重要的是，它更具可读性。现在，如果你需要舍入而不是“总是趋近于0”，你将需要稍微复杂一些的代码。

简单地说:

double d = 394.000;
long l = d * 1L;

．.． 这是舍入法，不会截断。赶紧在Java API手册里查了一下:

double d = 1234.56;
long x = Math.round(d); //1235

如果你强烈怀疑DOUBLE实际上是LONG，并且你想

1)将其EXACT值处理为LONG

2)当它不是LONG时抛出错误

你可以尝试这样做:

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是Double的子集，所以你可能会得到一些奇怪的结果，如果你不知情地尝试转换一个在Long范围之外的Double:

@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());
}

(new Double(d)). longvalue()在内部只执行强制转换，因此没有理由创建Double对象。