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Robot Framework:在Windows上以参数启动进程?

我是Robot Framework的新手,我找不到在Windows上运行带有参数的过程的方法。我相当确定我不了解文档,但是有一种简单的方法可以做到这一点...

好的,假设我可以使用以下命令启动程序:

c:\myappdir>MyApp.exe /I ..\params\myAppParams.bin 
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如何在RF中做到这一点?

任何帮助将不胜感激。非常感谢你 :)

编辑1:

这是我的一段代码:

| *Setting*            | *Value*
| Resource             | compilationResource.robot 
#(Process lib is included in compilationResource)

#I removed the "|" for readability
...
TEST1
...
  ${REPLAYEXEDIR}=  get_replay_exe_dir #from a custom lib included in compilationResource
  ${EXEFULLPATH}= Join Path  ${WORKSPACEDIR}  ${REPLAYEXEDIR}  SDataProc.exe
  Should Exist  ${EXEFULLPATH}
  ${REPLAYLOGPATH}=  Join Path  ${WORKSPACEDIR}  ReplayLog.log
  ${REPLAYFILEPATH}=  Join Path  ${WORKSPACEDIR}  params  params.bin
  Should Exist  ${REPLAYFILEPATH}

  Start Process  ${EXEFULLPATH}  stderr=${REPLAYLOGPATH}  stdout=${REPLAYLOGPATH}  alias=replayjob
  Process Should Be Running  replayjob
  Terminate Process  replayjob …
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