Created
June 14, 2013 16:39
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Examining __call__ of function objects - it's just a "method-wrapper," an object that wraps what is internally a method. I'm guessing the __call__ method is dynamically created when we ask for it, and there must be a short-circuit where function objects actually can get called with ().
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| >>> def foo(): return 1 | |
| >>> foo | |
| <function foo at 0x10c5af488> | |
| >>> foo.__call__ | |
| <method-wrapper '__call__' of function object at 0x10c5af488> | |
| >>> foo.__call__.__call__ | |
| <method-wrapper '__call__' of method-wrapper object at 0x10c562c50> | |
| >>> foo.__call__.__call__.__call__ | |
| <method-wrapper '__call__' of method-wrapper object at 0x10c64c8d0> | |
| >>> foo.__call__.__call__.__call__() | |
| 1 |
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