On Tue, 21 Dec 2021 23:01:20 GMT, Chris Plummer <cjplum...@openjdk.org> wrote:
>> src/jdk.hotspot.agent/share/classes/sun/jvm/hotspot/CommandProcessor.java
>> line 560:
>>
>>> 558: } else {
>>> 559: out.println();
>>> 560: }
>>
>> I wonder if it is possible to simplify the above fragment at 549-560 with
>> something like:
>> if (verbose) {
>> // If we know what this is a pointer to, then print
>> additional information.
>> PointerLocation loc = PointerFinder.find(val);
>> if (!loc.isUnknown()) {
>> out.print(" ");
>> loc.printOn(out, false, false); // make it
>> print without line end or ignore there is an extra end?
>> }
>> }
>> out.println();
>
> `printOn()` always prints a newline. It's hard to make it not do that since
> it relies on other `print` APIs, some of which print newlines and some of
> which don't (in which case `printOn()` will print a newline). So I can't
> easily just create a `printOn()` variant that doesn't print newlines, and
> printing two newlines would introduce a bunch of blank lines in the output,
> which wouldn't look good.
Okay, thanks.
-------------
PR: https://git.openjdk.java.net/jdk/pull/6902
