Why is a bareword following the concatenation operator a string in Perl?-CodePudding

Why does Perl (5.30.3 but also no feature ':all') assume a bareword following the concatenation operator . to be a string? For example:

use v5.30;   # or: no feature ':all'; 
use strict;
use warnings;
my $k1 = 'a';
my $k2 = 'b';
print STDOUT $k1 . k2 . "\n";

prints out ak2 (ab was intended), and raises no exceptions about barewords being not allowed here. This is a bug in my script, though not necessarily in Perl.

According to perlglossary, a bareword is a "word sufficiently ambiguous to be deemed illegal under use strict 'subs'." However, since a scalar named $k2 was defined before k2, I'd have thought k2 to be ambiguous enough.

Curiously, an exception is raised with a second bareword:

use v5.30;   # or: no feature ':all';
use strict;
use warnings;
my $k1 = 'a';
my $k2 = 'b';
my $k3 = 'c';
print STDOUT $k1 . k2 . k3  . "\n";

And:

Bareword "k3" not allowed while "strict subs" in use at mwe.pl line 7.
Execution of mwe.pl aborted due to compilation errors.

k3 is not allowed, but k2 is. Why?

CodePudding user response：

It's a bug introduced in 5.28.0 and fixed in 5.32.0.

A newly-added compile-time optimisation of (usually multiple) string concatenations was skipping the bareword check.