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# Help with Operator Precedence

Can someone explain to me why the OUTPUT of this code is "No"? x = 4 y = 2 if not 1+1 == y or x == 4 and 7 == 8 print("Yes") elif x>y print("No")

3 Answers

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because first if condition is false
if (not 1+1 == y) or ((x == 4) and (7 == 8))

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In canonical form the if-statemet looks like this:
(not ((1+1) ==y)) or ((x==4) and (7==8))
the priority of operators is as following:
+ == not and or.
so left part gives us:
(1+1) = 2
2 == y -> True (cuz y=2)
not (true) -> False
and right part gives us:
(x==4) -> True (cuz x=4)
(7==8) ->False (obvious)
(True and False) - > False (logical AND)
now we have both left and right parts False.
and finally:
(False) or (False) -> False (logiacal OR)
if-satement failed to succeed, so we go on with elif statement, which is True per conditions.

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Thank you for the help!