Groovy 3.0.19中picocli requiredOptionMarker类型转换报错求助

问题现象 #

脚本在Groovy 3.0.19中运行失败，但在Groovy 4.0.15中可正常运行。

背景 #

尝试遵循picocli文档中Groovy脚本相关说明，其中@Command注解的requiredOptionMarker属性要求接收char类型值。

报错场景 #

当为requiredOptionMarker赋值'*'时，编译失败：

@Command(name = "sampleScript",
        mixinStandardHelpOptions = true,
        requiredOptionMarker = '*')

报错信息如下：

Attribute 'requiredOptionMarker' should have type 'java.lang.Character'; but found type 'java.lang.String' in @picocli.CommandLine$Command
 @ line 63, column 32.
           requiredOptionMarker = '*',
                                  ^

1 error

已尝试的解决方法 #

• 尝试过(char) '*'、'*' as char等类型转换方式，均无效
• 找到相关Groovy问题工单：GROOVY-3278
• 使用((char)'*')赋值时，仍报错：
  代码示例：

  sortOptions = false,
      requiredOptionMarker = ((char)'*') //---->>>>>> This DOES"T WORK
  

  报错信息：

  >groovy sample-picocli.groovy
  
  org.codehaus.groovy.control.MultipleCompilationErrorsException: startup failed:
  C:\ws\sample-picocli.groovy: 35: Expected '(char) *' to be an inline constant of type char in @picocli.CommandLine$Command
   @ line 35, column 32.
             requiredOptionMarker = ((char)'*') //---->>>>>> This DOES"T WORK
                                    ^
  
  C:\ws\picocli\sample-picocli.groovy: 35: Attribute 'requiredOptionMarker' should have type 'java.lang.Character'; but found type 'java.lang.Object' in @picocli.CommandLine$Command
   @ line 35, column 32.
             requiredOptionMarker = ((char)'*') //---->>>>>> This DOES"T WORK
                                    ^
  
  C:\ws\picocli\sample-picocli.groovy: 35: Expected '(char) *' to be an inline constant of type char in @picocli.CommandLine$Command
   @ line 35, column 32.
             requiredOptionMarker = ((char)'*') //---->>>>>> This DOES"T WORK
                                    ^
  
  C:\ws\picocli\sample-picocli.groovy: 35: Attribute 'requiredOptionMarker' should have type 'java.lang.Character'; but found type 'java.lang.Object' in @picocli.CommandLine$Command
   @ line 35, column 32.
             requiredOptionMarker = ((char)'*') //---->>>>>> This DOES"T WORK
                                    ^
  
  4 errors
  

内容的提问来源于stack exchange，提问作者Asif Kamran Malick