> (int)(tokens.countTokens() / 0.75f) + 1 > (tokens.countTokens() * 4 + 2) / 3
if you want A * X / Y, rounded up, you can use (A * X - 1 ) / Y + 1 eg where X/Y = 4/3, 0 => 0 1 => 2 2 => 3 3 => 4 4 => 6 ... Mark On Tue, May 5, 2020 at 9:48 AM Peter Levart <peter.lev...@gmail.com> wrote: > Hi Naoto, > > On 4/30/20 12:18 AM, naoto.s...@oracle.com wrote: > > Hello, > > > > Please review this small fix to the following issue: > > > > https://bugs.openjdk.java.net/browse/JDK-8244152 > > > > The proposed changeset is located at: > > > > https://cr.openjdk.java.net/~naoto/8244152/webrev.00/ > > > > The hash map used there didn't have initial capacity, even though the > > exact numbers are known. > > > Well, it has to be calculated 1st (countTokens), but I guess this pays > off when HashSet (the backing HashMap) does not have to be rehashed then. > > The expression you use: > > Math.max((int)(tokens.countTokens() / 0.75f) + 1, 16) > > ...has a minimum value of 16. Why is that? 16 is just HashMap's default > initialCapacity if not specified explicitly. But if you only want to > store say 1 entry in the map, you can specify 2 as initialCapacity and > HashMap will happily work for such case without resizing. > > > So you could just use: > > (int)(tokens.countTokens() / 0.75f) + 1 > > And even this expression is sometimes overshooting the minimal required > value by 1 (when # of tokens is "exact" multiple of 0.75f, say 6). I > think the following could be used to optimally pre-size the HashMap with > default load factor 0.75: > > (tokens.countTokens() * 4 + 2) / 3 > > > Regards, Peter > > > > > Naoto >
