Question

我正在使用一个接受~~重写方法的类~~ 我的意思是可选参数签名（不确定在这种情况下是否重要，但也许）

当我从IRB调用它时，它按预期工作，例如，它接受参数

（用[过滤]过滤命名空间和密码，以保密秘密，我的公司很开心）

jruby-1.5.0 > require 'java'
 => true 

jruby-1.5.0 > Dir.glob('lib/java/*.jar').each{|jar| require jar}
 => ["lib/java/[filtered].jar", "lib/java/[filtered].jar", "lib/java/[filtered].jar"] 

jruby-1.5.0 > import "[filtered].His351n1"
 => Java::[filtered]::His351n1 

jruby-1.5.0 > broker = [filtered].Broker.new('[filtered]', '[filtered]')
 => #<Java::[filtered]::Broker:0x4c4936f3> 

jruby-1.5.0 > rpc = "[filtered]"
 => "[filtered]" 

jruby-1.5.0 > his = His351n1.new(broker, rpc)
 => #<Java::[filtered]::His351n1:0x7fb6a1c4>

这是我的规范和匹配代码

before(:each) do 
  @base = Legacy::Base.new
end

it "should create a valid his351n1 object" do
  his = @base.create_his351n1
  puts his.inpsect
end  

# from within Legacy::Base
def create_his351n1
  his = His351n1.new(build_broker, rpc)
end

最后，在对His351n1.new

的调用中失败的错误

1)
ArgumentError in 'Legacy::Base should create a valid his351n1 object'
wrong # of arguments(2 for 0)

为了使问题复杂化，在irb上，这显然也是有效的：

jruby-1.5.0 > his = His351n1.new
 => #<Java::[filtered]::His351n1:0x5ad3c69c>

此外，这里是重写的java方法

public His351n1() {
    super();
}

public His351n1(Broker broker) {
    this(broker, DEFAULT_SERVER);
}

public His351n1(BrokerService bs) {
    this(bs.getBroker(), bs.toString());
}

public His351n1(Broker broker, String serverAddr) {
    super(broker, serverAddr, "string", true);
}

public His351n1(final Broker broker, final String serverAddr, final String library)
{
    super(broker, serverAddr, library, true);
}

Answer 1

似乎你必须在对象的实例化中要求整个命名空间：

his = Java::[filtered_namesapce]::His351n1.new(build_broker, rpc)

jruby：java类接受IRB中的参数，但在从类调用时返回错误

1 个答案: