如何从Perl中的模块导出名为’import’的函数？

现在OP澄清了他的需求,这确实应该以类似于Exported的方式完成,确切地说,通过全局赋值将子引用注入调用者的命名空间.例：

###############################################

package ImportRenamer; 
use strict;
sub import_me {
   print "I am a cool importern";
}

sub import { 
  my ($callpkg)=caller(0);
  print "Setting ${callpkg}::import to ImportRenamer::import_men"; 
  no strict "refs";
  *{$callpkg."::import"} = &;ImportRenamer::import_me; # Work happens here!!!
  use strict "refs";
}
1;

###############################################

package My; 
use strict;
use ImportRenamer; 
1;

###############################################

package My2; 
use strict;
use ImportRenamer; 
1;

###############################################

而且测试：

> perl -e '{  package main; use My; use My2; 1;}'
Setting My::import to ImportRenamer::import_me
I am a cool importer
Setting My2::import to ImportRenamer::import_me
I am a cool importer

原始答案

除了调用导入方法“import”之外,您不需要做任何特殊操作.使用已经调用import(),参见perldoc use：

use Module LIST

Imports some semantics into the current package from the named module,
generally by aliasing certain subroutine or variable names into your package.

It is exactly equivalent to:

BEGIN { require Module; Module->import( LIST ); }