我正在尝试用 Perl6 做一些 OOP 并且在角色方面遇到了一些麻烦。我试图以与 Java 接口类似的方式使用它们,在那里我只有方法签名,必须由任何扮演该角色的类实现。我正在使用带有类型参数并返回的存根方法。
我注意到类型签名没有被强制执行,只有方法的名称。
示例脚本:
#!/usr/bin/env perl6
use v6;
role MyRole {
method intAdder( Int $a, Int $b --> Int ) { ... }
}
# this does the role and the method signature matches
class MyClass1 does MyRole {
method intAdder( Int $a, Int $b --> Int ) { return $a+$b }
}
# this does the role and the method signature does NOT match
# why is this allowed?
class MyClass2 does MyRole {
method intAdder( Str $a --> Str ) { return "Hello, $a." }
}
# this does the role and the method name does not match and gives an error as expected:
# Method 'intAdder' must be implemented by MyClass3 because it is required by roles: MyRole.
#
# class MyClass3 does MyRole {
# method adder( Int $a, Int $b --> Int ) { return $a+$b }
# }
sub MAIN() {
my $object1 = MyClass1.new;
my $object2 = MyClass2.new;
say $object1.intAdder: 40, 2;
say $object2.intAdder: 'world';
}
# output:
# 42
# Hello, world.
我已经阅读了官方文档中的面向对象页面,但找不到一种方法来做我想做的事......我也在尝试应用一种 Java 方式来思考 OOP 和打字,也许有不同的,更多Perl6ish 做我想做的事...