On Wed, 25 May 2022 13:28:52 GMT, Jorn Vernee <jver...@openjdk.org> wrote:
>> src/java.base/share/classes/java/lang/invoke/LambdaFormEditor.java line 239:
>>
>>> 237: for (int i = 0; i < b23456.length; i++) {
>>> 238: int b = b23456[i] & 0xFF;
>>> 239: bitset |= b;
>>
>> Looks like `b` is always truncated. I wonder what happens if the ints in
>> this array are larger than a byte (which seems to be possible in e.g. the
>> case of argument positions). Some higher order bits might be dropped, but
>> the resulting `b` might only have the least significant 4 bits set.
>>
>> I think the untruncated value should be used to compute the bitset? `butset
>> |= b23456[i]`? Then the `inRange` check should reject that case.
>
> Maybe not... argument positions should fit in a byte as well. But, maybe
> there are other problematic cases? Or are the ints guaranteed to fit in a
> byte?
Maybe an `assert b == b23456[i]` would be nice here.
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PR: https://git.openjdk.java.net/jdk/pull/8881
