Why does Math.round return a long but Math.floor return a double?

floor would have been chosen to match the standard c routine in math.h (rint, mentioned in another answer, is also present in that library, and returns a double, as in java).

but round was not a standard function in c at that time (it's not mentioned in C89 - c identifiers and standards; c99 does define round and it returns a double, as you would expect). it's normal for language designers to "borrow" ideas, so maybe it comes from some other language? fortran 77 doesn't have a function of that name and i am not sure what else would have been used back then as a reference. perhaps vb - that does have Round but, unfortunately for this theory, it returns a double (php too). interestingly, perl deliberately avoids defining round.

[update: hmmm. looks like smalltalk returns integers. i don't know enough about smalltalk to know if that is correct and/or general, and the method is called rounded, but it might be the source. smalltalk did influence java in some ways (although more conceptually than in details).]

if it's not smalltalk, then we're left with the hypothesis that someone simply chose poorly (given the implicit conversions possible in java it seems to me that returning a double would have been more useful, since then it can be used both while converting types and when doing floating point calculations).

in other words: functions common to java and c tend to be consistent with the c library standard at the time; the rest seem to be arbitrary, but this particular wrinkle may have come from smalltalk.

I agree, that it is odd that Math.round(double) returns long. If large double values are cast to long (which is what Math.round implicitly does), Long.MAX_VALUE is returned. An alternative is using Math.rint() in order to avoid that. However, Math.rint() has a somewhat strange rounding behavior: ties are settled by rounding to the even integer, i.e. 4.5 is rounded down to 4.0 but 5.5 is rounded up to 6.0). Another alternative is to use Math.floor(x+0.5). But be aware that 1.5 is rounded to 2 while -1.5 is rounded to -1, not -2. Yet another alternative is to use Math.round, but only if the number is in the range between Long.MIN_VALUE and Long.MAX_VALUE. Double precision floating point values outside this range are integers anyhow.

Unfortunately, why Math.round() returns long is unknown. Somebody made that decision, and he probably never gave an interview to tell us why. My guess is, that Math.round was designed to provide a better way (i.e., with rounding) for converting doubles to longs.

Like everyone else here I also don't know the answer, but thought someone might find this useful. I noticed that if you want to round a double to an int without casting, you can use the two round implementations long round(double) and int round(float) together:

double d = something;
int i = Math.round(Math.round(d));