Help: trouble with "SOURCE UNAVAILABLE"

[samhld]

Overall, this is working for me but instead of getting the line of code that's being evaluated, I'm getting a constant SOURCE UNAVAILABLE message. It looks like this is called from here. Not sure why it's not a "source" though. Can you help? Thanks!

This is happening on a very basic function, say:

@pysnooper.snoop()
def hello(name):
     print(f"Hello, {name}")

Returns:

> hello('sam')
Source path:... <stdin>
Starting var:.. name = 'sam'
17:27:50.065066 call         1 SOURCE IS UNAVAILABLE
17:27:50.065288 line         3 SOURCE IS UNAVAILABLE
Hello, sam
17:27:50.065439 return       3 SOURCE IS UNAVAILABLE
Return value:.. None

[cool-RR]

What's the scenario? Where is your function defined? In a code file, in a shell, and if so what shell?

On Fri, Oct 18, 2019, 03:34 Sam Dillard notifications@github.com wrote:

Overall, this is working for me but instead of getting the line of code that's being evaluated, I'm getting a constant SOURCE UNAVAILABLE message. It looks like this is called from here https://github.com/cool-RR/PySnooper/blob/master/pysnooper/tracer.py#L89. Not sure why it's not a "source" though. Can you help? Thanks!

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[samhld]

@cool-RR it's defined in a python3 interactive shell. Sounds like maybe I don't understand what "source" means in this case?

[alexmojaki]

The source code isn't saved anywhere in a regular shell. If you write code that raises an exception you'll see that code doesn't show in the traceback either.

[cool-RR]

@alexmojaki is correct. If you were to suggest to python-ideas to start saving source code on code objects, I would support that suggestion, and that would make this feature possible. I think that they're unlikely to accept it, but let me know if you try.

[samhld]

thanks, guys! Clears that up.