Question

package a;
sub func {
print 1;
}
package main;
a::->func;

IMO足以拥有a::func，a->func。

a::->func;对我来说很奇怪，为什么Perl支持这种奇怪的语法？

Answer 1

引用chromatic在Modern Perl blog上关于该主题的优秀近期博客文章：“避免使用匿名解析歧义。”

为了说明为什么这样的语法有用，这里是一个从你的样本演变而来的例子：

package a;
our $fh;
use IO::File;
sub s {
    return $fh = IO::File->new();
}

package a::s;
sub binmode {
    print "BINMODE\n";
}

package main;
a::s->binmode; # does that mean a::s()->binmode ?
               # calling sub "s()" from package a; 
               # and then executing sub "open" of the returned object?
               # In other words, equivalent to $obj = a::s(); $obj->binmode();
               # Meaning, set the binmode on a newly created IO::File object?

a::s->binmode; # OR, does that mean "a::s"->binmode() ?
               # calling sub "binmode()" from package a::s; 
               # Meaning, print "BINMODE"

a::s::->binmode; # Not ambiguous - we KNOW it's the latter option - print "BINMODE"

Answer 2

a::是一个字符串文字，用于生成字符串a。一切都是这样的：

 a->func()    # Only if a doesn't exist as a function.
 "a"->func()
 'a'->func()
 a::->func()
 v97->func()
 chr(97)->func()

等

>perl -E"say('a', a, a::, v97, chr(97))"
aaaaa

为什么`a :: - ＆gt; func;`有效？

2 个答案: