Why does Math.ceil return a double?

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) double value 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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最佳答案

The range of double is greater than that of long. For example:

double x = Long.MAX_VALUE;
x = x * 1000;
x = Math.ceil(x);

What would you expect the last line to do if Math.ceil returned long?

Note that at very large values (positive or negative) the numbers end up being distributed very sparsely - so the next integer greater than integer x won't be x + 1 if you see what I mean.

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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.