考虑构建一个集合并测试成员资格的代码：

my $num_set = set( < 1 2 3 4 > );
say "set: ",$num_set.perl;
say "4 is in set: ",4 ∈ $num_set;
say "IntStr 4 is in set: ",IntStr.new(4,"Four") ∈ $num_set;
say "IntStr(4,...) is 4: ","Four") == 4;
say "5 is in set: ",5 ∈ $num_set;

一个直4不在集合,但IntStr版本是：

set: set(IntStr.new(4,"4"),IntStr.new(1,"1"),IntStr.new(2,"2"),IntStr.new(3,"3"))
4 is in set: False
IntStr 4 is in set: True
IntStr(4,...) is 4: True
5 is in set: False

我认为大多数人都不会期待这一点,但是∈docs并没有说明这个方法是如何工作的.如果我不使用报价单(即设置(1,2,3,4)),我没有这个问题.