为什么 Math.ceil 返回双精度？
当我调用 Math.ceil(5.2) 时，返回的是 double 6.0。我的自然倾向是认为 Math.ceil(double a) 会返回一个 long 。从文档中：
ceil(双a)返回最小（最接近负无穷大）
double值 不小于参数并且等于数学 整数。
但是，当结果是整数时，为什么要返回 double 而不是 long 呢？我认为理解其背后的原因可能会帮助我更好地理解 Java。它还可以帮助我弄清楚转换为 long 是否会给自己带来麻烦，例如 is
long b = (long)Math.ceil(a);
总是我认为应该的样子？我担心可能存在一些有问题的边界情况。
When I call Math.ceil(5.2) the return is the double 6.0. My natural inclination was to think that Math.ceil(double a) would return a long. From the documentation:
ceil(double a)Returns the smallest (closest to negative infinity)
doublevalue
that is not less than the argument and is equal to a mathematical
integer.
But why return a double rather than a long when the result is an integer? I think understanding the reason behind it might help me understand Java a bit better. It also might help me figure out if I'll get myself into trouble by casting to a long, e.g. is
long b = (long)Math.ceil(a);
always what I think it should be? I fear there could be some boundary cases that are problematic.
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double的范围大于long的范围。例如：如果
Math.ceil返回long，您希望最后一行做什么？请注意，在非常大的值（正数或负数）下，数字最终会分布得非常稀疏 - 因此，下一个大于整数
x的整数将不会是x + 1，如果你明白我的意思了。The range of
doubleis greater than that oflong. For example:What would you expect the last line to do if
Math.ceilreturnedlong?Note that at very large values (positive or negative) the numbers end up being distributed very sparsely - so the next integer greater than integer
xwon't bex + 1if you see what I mean.双精度数可以大于
Long.MAX_VALUE。如果您对这样的值调用 Math.ceil() ，您会期望返回相同的值。但是，如果它返回 long，则该值将不正确。A double can be larger than
Long.MAX_VALUE. If you callMath.ceil()on such a value you would expect to return the same value. However if it returned a long, the value would be incorrect.