CODEWITHSUNDEEP
pythonscopeexec
Tuom L.
Tuom L.

Reputation: 182

Python exec and __name__

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?

Upvotes: 6

Views: 2654

Answers (2)

Xion
Xion

Reputation: 22770

One solution is to provide the __name__ explicitly in your execution dict:

exec("print(__name__)", {'__name__': '__main__'})

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121366

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!

Upvotes: 6

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