Records are called „methods“ in JShell

[Lingo Coder]

Thanks Robert,
 
>  „...Can you do:
>  jshell --full-version...“
 
I did that. It says jshell 15-ea+31-1502 [1]

What version of Windows are you other guys trying this in? I'm on 
Windows 7.

> „...I'm guessing you may be using an older JDK...“

As I show in [1] and in earlier screen recordings, I completely
clear my $PATH to confirm that there is no other jshell.exe on
the $PATH.

I have not installed any JDKs with an installer or anything
like that. For that reason, I rule out anything registry-related. 

I do have an early access release of 14-Valhalla 
build 14-valhalla+4-55 on my workstation. But, that was 
installed by simply unzipping the archive into a 
jdk-14.L.world.Build.14-valhalla+4-55 folder. More importantly,
that build doesn't even have the records nor the sealed types
features.

I also have GA releases of OpenJDK 12 & 13 in their own folders.
But again, not only are they isolated well out of the way of
the $PATH I run jshell 15 in, they don't even have records and
sealed types functionality.

If _any_ other JDK were somehow in the $PATH, it would be
surprising if records and sealed types even worked _at all_;
let alone the misnaming as „methods“ issue Right? Please 
correct me if I'm wrong?


[1] https://imgur.com/cMsAl5f

-------- Original Message --------
Subject: Re:_Records_are_called_„methods“_in_JSh ell
From: Robert Field <robert.field at oracle.com>
Date: Tue, July 14, 2020 3:57 pm
To: kulla-dev at openjdk.java.net, jorn.vernee at oracle.com,
plugins at lingocoder.com

 I'm guessing you may be using an older JDK.  I get:
  jshell> public record ConstExpr(Expr e) implements Expr{}
 |  created record ConstExpr
 
 jshell> /drop ConstExpr
 |  dropped record ConstExpr
 
 
  Can you do:
  jshell --full-version
 
  Thanks,
 Robert