perl6 语法操作:如果不使用 $/则无法进行任何操作 [英] perl6 grammar actions: unable to make anything if not using $/
问题描述
我写了一个测试程序，现在看来如果不使用$/方法签名因为我必须在方法内部使用 .match ，我不能再做任何事情了.我做错了什么?
I wrote a test program, and now it seems that if I don't use $/ in a
method signature because I have to use .match inside the method, I can no long make anything. What did I do wrong?
另一个问题是，如果.match 设置了$/，并且$/ 是只读的，那么我不能有$/ 在包含 .match 语句的方法的签名中，并且我不能在方法内部有多个 .match 因为每个 .match 将尝试设置只读 $/.这样编程会很别扭.
A further question is that if .match sets $/, and $/ is read-only, then I cannot have $/ in the signature of a method that contains a .match statement, and I cannot have more than one .match inside the method because each .match will try to set the read-only $/. This will be very awkward to program.
这是里面只有一个.match语句的测试程序和结果:
Here is the test program with only one .match statement inside and the results:
grammar test {
regex TOP { <foo><bar> }
regex foo { :i s* foo s* }
regex bar { :i s bar s* }
}
class actTest {
method foo ($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 $m;
结果:
1 ｢Foo ｣
2 (Match)
3 Foo
4 (Match)
5 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
6 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
10 oo
11 F
12 ooF
1 ｢Foo｣
2 (Match)
3 Foo
4 (Match)
5 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
6 ｢Foo｣
0 => ｢F｣
1 => ｢oo｣
10 oo
11 F
12 ooF
8 ｢ bar｣
9 (Any)
Use of uninitialized value of type Any in string context.
Methods .^name, .perl, .gist, or .say can be used to stringify it to
something meaningful.
in method TOP at matchTest.pl line 28
7 ｢Foo bar｣
foo => ｢Foo｣
bar => ｢ bar｣
推荐答案
1) 如何在没有$/
的情况下makemake ... 只是 $/.make(...) 的快捷方式.
1) How to make without $/
make ... is just a shortcut for $/.make(...).
如果您想影响与 $/ 中存储的对象不同的 Match 对象，则必须直接使用方法形式，即在您的情况下 $x.make($result).
If you want to affect a different Match object than the one stored in $/, you have to use the method form directly, i.e. in your case $x.make($result).
默认情况下，$/ 绑定到一个普通的 item 容器(就像一个用 my 声明的变量)，即 not 只读.因此，在例程中多次使用 .match 方法应该没有任何问题.
By default, $/ is bound to a normal item container (like a variable declared with my), i.e. not read-only. So there shouldn't be any problem with using the .match method multiple times in a routine.
只有在例程签名中明确声明$/为参数时，$/才会直接绑定到Match对象传递给该例程(未包装在项目容器中)，因此在例程中将是只读的 - 因为这是正常签名绑定的工作方式.
It's only when you explicitly declare $/ as a parameter in a routine signature, that $/ will be bound directly to the Match object passed to that routine (not wrapped in an item container), and will thus be read-only inside the routine – because that's how normal signature binding works.
您可以使用 is copy trait 来覆盖普通的只读参数绑定，并强制 $/ 成为例程内的可变项目容器:
You could use the is copy trait to override the normal read-only parameter binding, and force $/ to be a mutable item container inside the routine:
method foo ($/ is copy) { ... }
这样，在例程中使用 .match 就可以了，并且会在 $/ 中存储一个新的 Match 对象.但是，您将无法再访问传递给例程的原始 Match 对象，因此无法使用 make 影响它.因此，对于需要使用 .match 的操作方法，像您一样使用自定义参数名称是可行的方法.
This way, using .match inside the routine would work, and would store a new Match object in $/. But then you wouldn't have access to the original Match object passed to the routine anymore, and thus couldn't affect it with make. So for an action method that needs to use .match, using a custom parameter name like you did is the way to go.
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