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I've just installed VSCode (version 1.30.2) and am setting it up to use with Python. I want to enable debugging. The default launch.json contains this entry:

{
    "name": "Python: Current File (Integrated Terminal)",
    "type": "python",
    "request": "launch",
    "program": "${file}",
    "console": "integratedTerminal"
}

I have modified it to also include the debugOptions:

{
    "name": "Python: Debug",
    "type": "python",
    "request": "launch",
    "program": "${file}",
    "debugOptions": [
        "WaitOnAbnormalExit",
        "WaitOnNormalExit",
        "RedirectOutput"
    ]
}

However, VSCode warns me about debugOptions with the following warning:

Property debugOptions is not allowed.

I've seen this question but the answer doesn't seem to actually change the warning. Additionally, I don't seem to need the line that the answer indicates I need.

Debugging works with the configuration I have, but I want to remove the warning that VSCode is showing on "debugOptions". How can I do so?

NewGuy
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  • The setting `debugOptions` has been removed from the debug configuration file since around the august 2018 release of the Python Extension. As a maintainer of the extension I'd like to understand why you would like to use that setting? – Don Jan 23 '19 at 07:52
  • @Don I'm new with VSCode, so I'm not entirely sure if it's the appropriate option to use. I was following the documentation [here](https://code.visualstudio.com/docs/python/debugging) and found the parameter and [here](https://donjayamanne.github.io/pythonVSCodeDocs/docs/debugging_library-functions/) for possible values (also found in this [answer](https://stackoverflow.com/a/36868320/) ) – NewGuy Jan 23 '19 at 19:49

0 Answers0