def foo1():
return True == 3 in [1,2,3]
def foo2():
return 3 in [1,2,3] == True
def foo3():
return True == ( 3 in [1,2,3] )
def foo4():
return ( 3 in [1,2,3] ) == True
foo1() => False
foo2() => False
foo3() => True
foo4() => True
import dis
dis.dis(foo1)
2 0 LOAD_CONST 1 (True)
2 LOAD_CONST 2 (3)
4 DUP_TOP
6 ROT_THREE
8 COMPARE_OP 2 (==)
10 JUMP_IF_FALSE_OR_POP 18
12 LOAD_CONST 5 ((1, 2, 3))
14 COMPARE_OP 6 (in)
16 RETURN_VALUE
>> 18 ROT_TWO
20 POP_TOP
22 RETURN_VALUE
dis.dis( foo2 )
2 0 LOAD_CONST 1 (3)
2 LOAD_CONST 2 (1)
4 LOAD_CONST 3 (2)
6 LOAD_CONST 1 (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 2 (==)
22 RETURN_VALUE
>> 24 ROT_TWO
26 POP_TOP
28 RETURN_VALUE
>>> dis.dis( foo3 )
2 0 LOAD_CONST 1 (True)
2 LOAD_CONST 2 (3)
4 LOAD_CONST 5 ((1, 2, 3))
6 COMPARE_OP 6 (in)
8 COMPARE_OP 2 (==)
10 RETURN_VALUE
>>> dis.dis( foo4 )
2 0 LOAD_CONST 1 (3)
2 LOAD_CONST 5 ((1, 2, 3))
4 COMPARE_OP 6 (in)
6 LOAD_CONST 4 (True)
8 COMPARE_OP 2 (==)
10 RETURN_VALUE
From the disassemble code I can see the foo1() should return False since it won't execute the in comparison and hence False.
But what happens to foo2(), why does this return False?
Whereas foo3() and foo4() works perfectly as we can be confirmed from the disassembled code.
Ref: https://docs.python.org/2/library/dis.html
Python2.7
I've seen similar behavior python3+; the disassembled code is different but logic seems to be same in the disassembled code.
Python operator precedence: https://docs.python.org/2/reference/expressions.html#operator-precedence
Operator precedence for== and in is same but it doesn't match the expectation.
