Nick Coghlan added the comment:
The origin of the ".0" parameter name is the compiler's implicit symbol
definition - we use the "." prefix for those to ensure we don't accidentally
collide with actual identifiers used in the code.
We can see that via dis.dis:
>>> def make_set():
... return {z*z for z in range(5)}
...
>>> import dis
>>> dis.dis(make_set)
2 0 LOAD_CONST 1 (<code object <setcomp> at
0x7f54c5b2cd20, file "<stdin>", line 2>)
3 LOAD_CONST 2 ('make_set.<locals>.<setcomp>')
6 MAKE_FUNCTION 0
9 LOAD_GLOBAL 0 (range)
12 LOAD_CONST 3 (5)
15 CALL_FUNCTION 1 (1 positional, 0 keyword pair)
18 GET_ITER
19 CALL_FUNCTION 1 (1 positional, 0 keyword pair)
22 RETURN_VALUE
>>> dis.dis(make_set.__code__.co_consts[1])
2 0 BUILD_SET 0
3 LOAD_FAST 0 (.0)
>> 6 FOR_ITER 16 (to 25)
9 STORE_FAST 1 (z)
12 LOAD_FAST 1 (z)
15 LOAD_FAST 1 (z)
18 BINARY_MULTIPLY
19 SET_ADD 2
22 JUMP_ABSOLUTE 6
>> 25 RETURN_VALUE
It isn't just set comprehensions that will do this - it's all the implicit
nested functions that accept a positional argument (list/dict/set
comprehensions, generator expressions)
I chatted to Brett Cannon about it, and we agree inspect should handle the
implicit functions generated by the compiler, even if those signatures can't be
expressed in normal Python code. The fact those may be emitted by the affected
inspect functions can then be documented as a CPython implementation detail,
rather than as a language level behaviour.
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue19611>
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