我在 Rakudo 2016.07 中运行良好的操作类中有以下方法，但我刚刚安装了 2016.11，现在新的 Rakudo 说我的方法试图分配给只读变量，我只是没有看到问题：
method ptName ($/) {
my $nameStr = $/.Str, my $lastName, my $firstName;
my $newMatch # this is line 182;
# Cannot assign to a readonly variable or a value
= $nameStr.match(/ \" (<alpha>+) .*? \, \s* (<alpha>+) .*? \" /);
$lastName = $newMatch[0];
$firstName = $newMatch[1];
make "$lastName $firstName";
}
整个错误信息是
Cannot assign to a readonly variable or a value
in method ptName at /home/lisprog/Binary/grammar.pl line 182
in regex ptName at /home/lisprog/Binary/grammar.pl line 151
in regex TOP at /home/lisprog/Binary/grammar.pl line 137
in block <unit> at /home/lisprog/Binary/grammar.pl line 217
什么语言规范发生了变化？请帮忙。谢谢。
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谢谢雷夫，克里斯托夫，ZZ！我不知道如何使用正确的格式添加长评论。所以，我在自己的帖子中添加评论。
我写了一个测试程序，现在看来，如果我在方法签名中不使用（$/），因为我必须在方法内部使用 .match，我将无法再做任何事情。我做错了什么？这是测试程序和结果：
测试程序：
grammar test {
regex TOP { <foo><bar> }
regex foo { :i \s* foo \s* }
regex bar { :i \s bar \s* }
}
class actTest {
method foo ($x) { # program fails if I use $/ in signature
print "1 "; say $x; # how to combine the 2 and show $x as match?
print "2 "; say $x.WHAT;
my $newStr = $x.Str;
print "3 "; say $newStr;
my $newMatch
= $newStr.match(/:i(f)(oo)/); # adverb cannot be outside?
print "4 "; say $newMatch.WHAT;
print "5 "; say $newMatch;
print "6 "; say $/;
my $oo = $newMatch[1].Str;
print "10 "; say $oo;
my $f = $newMatch[0].Str;
print "11 "; say $f;
my $result = $oo ~ $f;
print "12 "; say $result;
make $result; # now I cannot make anything; huh???
}
method TOP ($/) {
print "8 "; say $<bar>;
print "9 "; say $<foo>.made; # failed, method 'foo' makes nothing
make $<bar> ~ $<foo>.made;
}
}
my $m = test.parse("Foo bar", actions => actTest.new);
print "7 "; say $m;
结果：
1 ｢Foo ｣
2 (Match)
3 Foo
4 (Match)
5 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
6 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
10 oo
11 F
12 ooF
1 ｢Foo｣
2 (Match)
3 Foo
4 (Match)
5 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
6 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
10 oo
11 F
12 ooF
8 ｢ bar｣
9 (Any)
Use of uninitialized value of type Any in string context.
Methods .^name, .perl, .gist, or .say can be used to stringify it to
something meaningful.
in method TOP at matchTest.pl line 28
7 ｢Foo bar｣
foo => ｢Foo｣
bar => ｢ bar｣