perl - 为什么“a::-> func;”有效？

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

IMO有a::func，a->func就足够了。

a::->func;在我看来很奇怪，为什么Perl支持这种奇怪的语法？

最佳答案

在Modern Perl blog上引用chrom近期关于该主题的出色博客文章：“避免裸词解析歧义。”

为了说明为什么使用这种语法，下面是从您的示例演变而来的示例：

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"

关于perl - 为什么“a::-> func;”有效？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/7303806/