In REPL, user defined operators do not persist between lines operator

[ribbon-otter]

The Problem

Normal user defined subs persist between calls. If I declare sub a {}, a()works as the next command. This does not work for operators. If a is a infix, the next command will not have a definition for a at all.

Expected Behavior

For custom operators to last between commands. Since other state lasts between commands like this, I would expect for operators to last as well.

Actual Behavior

Custom operators are forgotten for each new command.

Steps to Reproduce

user[~]$perl6
To exit type 'exit' or '^D'
> sub infix:<bb> {$^a + $^b}; 2 bb 2
4
> sub infix:<bb> {$^a + $^b};
sub infix:<bb> ($a, $b) { #`(Sub+{Precedence}|94152335641936) ... }
> 2 bb 2
===SORRY!=== Error while compiling:
Two terms in a row
------> 2⏏ bb 2
    expecting any of:
        infix
        infix stopper
        statement end
        statement modifier
        statement modifier loop

Environment

• Operating system: Debian 6.3
• Compiler version (perl6 -v):

This is Rakudo version 2018.06 built on MoarVM version 2018.06

implementing Perl 6.c.

I also reproduced the same problem on a few weeks old recent development build which output this for perl -v:

This is Rakudo version 2018.06-449-gec366c30d built on MoarVM version 2018.06-425-g0daf99bcb

implementing Perl 6.c.

[zoffixznet]

Same as RT#131900. Grammar mutations don't persist in REPL.

[tinmarino]

We fixed this issue in p6-jupyter-kernel:

1. Saving the $?LANG from inside the REPL
2. Updating $!compiler : $compiler.parsegrammar($lang.slang_grammar('MAIN')) at each loop.

Would I make the same PR here ? Extracting information from the "sandbox" with a dynamic variable also ? We still have troubles with pragma persistence, details in this issue.