正则表达式 – 在Perl中,有没有${foo [bar]}可能是语法上有效的

I am reading through O’Reilly’s Programming Perl,3^rd Edition,and the text states that instead of using the ambiguous search pattern /$foo[bar]/,one should instead use /${foo[bar]}/ so that Perl doesn’t mistake [bar] for a character class. Am I missing something,or are both of these statements syntactically incorrect due to the fact that they are trying to index into an array using a bareword string?

是的,你错过了一些东西：bar可以是一个函数调用：

$perl -Mstrict -E 'sub bar() { 0 } say "foo" =~ /$ARGV[bar]/ || "FAIL"' foo
FAIL

$perl -Mstrict -E 'sub bar() { 0 } say "foo" =~ /${ARGV[bar]}/ || "FAIL"' foo
1

$perl -MO=Deparse -Mstrict -E 'sub bar() { 0 } say "foo" =~ /${ARGV[bar]}/ || "FAIL"' foo
sub bar () { 0 }
use strict 'refs';
BEGIN {
    $^H{'feature_unicode'} = q(1);
    $^H{'feature_say'} = q(1);
    $^H{'feature_state'} = q(1);
    $^H{'feature_switch'} = q(1);
}
say 'foo' =~ /$ARGV[0]/ || 'FAIL';
-e syntax OK

确切的引用,来自Programming Perl,4^th edition的第73页,是：

Within search patterns,which also undergo double-quotish interpolation,there
is an unfortunate ambiguity: is /$foo[bar]/ to be interpreted as /${foo}[bar]/
(where [bar] is a character class for the regular expression),or as /${foo[bar]}/
(where [bar] is the subscript to array @foo)? If @foo doesn’t otherwise exist,it’s
obviously a character class. If @foo exists,Perl takes a good guess about [bar] and is almost always right. If it does guess wrong,or if you’re just plain paranoid,you can force the correct interpretation with braces as shown earlier. Even if you’re merely prudent,it’s probably not a bad idea.