When I run:
exec("print(__name__)")
it prints __main__.
But when I run:
exec("print __name__", {})
it prints builtins.
How to make the second example to also print __main__?
What I try to achieve is to run a piece of code with exec() so that from the perspective of the it looks like it was run from command line.
I would like to tun the code with clean scope but the second example breaks the code relying on if __name__ == "__main__". How to fix this?
You could use imp.load_module instead:
import imp
with open(mainfile) as src:
imp.load_module('__main__', src, mainfile, (".py", "r", imp.PY_SOURCE))
This imports the file as the __main__ module, executing it.
Note that it takes an actual file object when the type is set to imp.PY_SOURCE, so you'd need to create a temporary file for this to work if your source code comes from somewhere other than a file.
Otherwise, can always set __name__ manually:
>>> src = '''\
... if __name__ == '__main__': print('Main!')
... else: print('Damn', __name__)
... '''
>>> exec(src)
Main!
>>> exec(src, {})
Damn builtins
>>> exec(src, {'__name__':'__main__'})
Main!
import) the file - does it import or execute the file? Is there a difference?
__name__ == '__main__' works implies you are executing modules here.src string? And I have to ask similar question as for the other answer - is in {'__name__':'__main__'} enough to make src believe it was executed from command line? I mean are there any traps I should know about? If it is enough though that would make it for a perfect solution, thanks in any case...__name__ is a module global, in modules globals and locals are the same thing so you should not supply a locals mapping. Testing for __name__ is the defacto way to test if this is the main script, setting it is enough.One solution is to provide the __name__ explicitly in your execution dict:
exec("print(__name__)", {'__name__': '__main__'})
__doc__ but is this fix essential to making it look like code from command line?