Perl6 String Concatenation Hangs - concatenation

Question

I'm trying to concatenate a string with Perl6 thus:
my $cmd = "databricks jobs --job-id 37 --notebook-params '\{";
put $cmd;
$cmd ,= "\"directory\": \"$s3-dir\",";
put $cmd;
However, the output hangs after the ,= operator. I'm assuming that this works the same way from the Perl .= operator.
https://docs.perl6.org/language/operators
Why is this job hanging? how can I properly concatenate the string?

Answer 1

There is no ,= or ~= operators in Perl6.
Those are instances of the = meta operator combined with another infix operator.
# these are all functionally equivalent:
$a = $a ~ 'a';
$a ~= 'a';
$a [~]= 'a';
$a [&[~]]= 'a';
$a [&infix:<~>]= 'a';
$a = infix:<~> $a, 'a'; # use an operator as a subroutine
There is a history of C-like languages having operators like +=.
It would get a little tiring having to define new operators like that for every infix operator.
In Perl6 you can also easily define new operators.
So it has = as a meta-operator that will automatically work with all infix operators.
sub infix:<foo> (\l,\r){…}
$a = $a foo 3;
$a foo= 3;
$a [foo]= 3;
If you want to use an operator like +=, just look for the base operator that matches what you want and add =.
If you want to do string concatenation, the base operator is the infix ~ operator.
(Which looks a lot like the string coercion prefix operator ~.)
$a = $a ~ 'a';
$a ~= 'a';
If you want to do Set difference, the base operator is (-).
$a = $a (-) 3;
$a (-)= 3;
You can add any number of [] surrounding an infix operator.
(It needs space before it so that it isn't confused with postcircumfix [])
$a - 3;
$a [-] 3;
$a [[-]] 3;
$a [[[-]]] 3;
Which can be useful to make sure that meta operators combine the way you want them too.
$a -= 3;
$a [-]= 3;
3 R-= $a; # $a = $a - 3;
$a [R-]= 3; # $a = 3 - $a;
3 R[-=] $a; # $a = $a - 3;
3 R[[R-]=] $a; # $a = 3 - $a;
This was extended so that [&…] where &… is the name of a function works as an infix operator.
sub bar (\l,\r){…}
# these are functionally identical.
$a = bar( $a, 3 );
$a = $a [&bar] 3;
$a [&bar]= 3;
When you used ,= you created a self-referential data structure.
(Note that say calls .gist, I added an extra .gist to be extra clear that I'm not printing a Str.)
my $c = 0;
# $c ,= 1;
$c = ($c,1);
say $c.gist;
# (\List_94195670785568 = (List_94195670785568 1))
say $c.WHICH;
# List|94195670785568
When you do something that coerces a List or Array into a Str, it follows the structure turning each part into a Str.
$c = ($c,…);
~$c;
# $c.Str
# | \___________________
# | \
# $c[0].Str ~ ' ' ~ $c[1].Str
# | \ \ \__________
# | | \ \__________________ 1.Str
# | | \
# | V \_____________
# | \
# $c[0].Str ~ ' ' ~ $c[1].Str
# | \ \ \__________
# | | \ \__________________ 1.Str
# | | \
# | V \_____________
# | \
# $c[0].Str ~ ' ' ~ $c[1].Str
# | \ \ \__________
# | | \ \__________________ 1.Str
# | | \
# | V \_____________
# | \
# …
It of course never finishes turning the first part into a Str.
The combination of , and = works safely with a Hash though.
(Which is what the docs show.)
my %c = a => 0;
%c ,= b => 1;
# %c = (%c, b => 1)
say %c.gist;
# {a => 0, b => 1}

Answer 2

There was some confusion in the manual.
my $cmd = "databricks jobs --job-id 37 --notebook-params '\{";
put $cmd;
$cmd ,= "\"directory\": \"$s3-dir\",";
put $cmd;
should be
my $cmd = "databricks jobs --job-id 37 --notebook-params '\{";
put $cmd;
$cmd ~= "\"directory\": \"$s3-dir\",";
put $cmd;
the proper way to concatenate an already declared variable in Perl6 is with ~=