(no title)
vanni | 5 years ago
In [1]: import dis
In [2]: def a():
...: return (1 in [1,2,3] is True)
In [3]: def b():
..: return ((1 in [1,2,3]) is True)
In [4]: a()
Out[4]: False
In [5]: b()
Out[5]: True
In [6]: dis.dis(a)
2 0 LOAD_CONST 1 (1)
2 LOAD_CONST 1 (1)
4 LOAD_CONST 2 (2)
6 LOAD_CONST 3 (3)
8 BUILD_LIST 3
10 DUP_TOP
12 ROT_THREE
14 COMPARE_OP 6 (in)
16 JUMP_IF_FALSE_OR_POP 24
18 LOAD_CONST 4 (True)
20 COMPARE_OP 8 (is)
22 RETURN_VALUE
>> 24 ROT_TWO
26 POP_TOP
28 RETURN_VALUE
In [7]: dis.dis(b)
2 0 LOAD_CONST 1 (1)
2 LOAD_CONST 5 ((1, 2, 3))
4 COMPARE_OP 6 (in)
6 LOAD_CONST 4 (True)
8 COMPARE_OP 8 (is)
10 RETURN_VALUE
Why?!
vanni|5 years ago
Maybe it is related to grammar, and not to compiler...
AST dump in CPython 3.6.9 (manually formatted):
vanni|5 years ago
zelphirkalt|5 years ago
vanni|5 years ago
1) "in" and "is" have same precedence, and group left to right (see https://docs.python.org/3/reference/expressions.html#operato...)
2) you'll have runtime error:
It seems related to CPython bytecode compiler implementation, the two functions are parsed in a different way, parentheses make the compiler go on a different path... but I'd like to understand why, without diving into CPython source code :) Anyone?morelisp|5 years ago