java 为什么 Math.ceil 返回双精度值？
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Why does Math.ceil return a double?
提问by PengOne
When I call Math.ceil(5.2)the return is the double6.0. My natural inclination was to think that Math.ceil(double a)would return a long. From the documentation:
当我打电话时Math.ceil(5.2)，返回的是double6.0. 我自然倾向于认为这Math.ceil(double a)会返回一个long. 从文档：
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.
ceil(double a)返回
double不小于参数且等于数学整数的最小（最接近负无穷大）值。
But why return a doublerather than a longwhen 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
但是为什么当结果是整数时返回 adouble而不是 along呢？我认为理解其背后的原因可能有助于我更好地理解 Java。它也可能帮助我弄清楚我是否会通过强制转换为 a 给自己带来麻烦long，例如是
long b = (long)Math.ceil(a);
long b = (long)Math.ceil(a);
always what I think it should be? I fear there could be some boundary cases that are problematic.
总是我认为它应该是什么？我担心可能会有一些有问题的边界案例。
采纳答案by Jon Skeet
The range of doubleis greater than that of long. For example:
的范围double大于的范围long。例如：
double x = Long.MAX_VALUE;
x = x * 1000;
x = Math.ceil(x);
What would you expect the last line to do if Math.ceilreturned long?
如果Math.ceil返回，您希望最后一行做long什么？
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 be x + 1if you see what I mean.
请注意，在非常大的值（正数或负数）下，数字最终会非常稀疏地分布 - 因此，如果您明白我的意思，则下一个大于整数的整数x将不会x + 1是。
回答by Peter Lawrey
A double can be larger than Long.MAX_VALUE. If you call Math.ceil()on such a value you would expect to return the same value. However if it returned a long, the value would be incorrect.
double 可以大于Long.MAX_VALUE. 如果您调用Math.ceil()这样的值，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。
