all() function not returning expected value in Python

Question

In Python,

a=[not(0)+1] 
print(all(a))

This returns False. But shouldn't it be True? Since, a=[2], and all integer values other than 0 are considered True. Thus, all() should return True right?

This has nothing to do with all - your question should be, why is not(0) + 1 false? You claim that a = [2] but if it were then you wouldn't have had to write not(0) + 1 to produce the issue. — kaya3
– kaya3, Commented Jan 30, 2021 at 12:01
The brackets are misleading; not is an operator, not a function, and has lower precedence than +, so you are effectively doing not ((0) + 1). — kaya3
– kaya3, Commented Jan 30, 2021 at 15:26

Aakash aggarwal · Accepted Answer · 2021-01-30 11:20:58Z

1

Yes, all() should return True if all the values in a list are true. In your case if you do

a = [2]
print(all(a))

this will return true. But when you are doing

a = [not(0) +1]

your list of a is becoming a = [False] as not(0) + 1 returns False.

a = [2]
a = [not(0) + 1]
print(a)
# a = [False]
print(all(a))
# False

Hence, this prints result as False

edited Jan 30, 2021 at 11:20

answered Jan 30, 2021 at 11:13

Aakash aggarwal

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3 Comments

Daniel Hao Over a year ago

not 0 == True (1) *Always!

Biswadeep Roy Over a year ago

Allright. But I still do not understand how not(0)+1 returns False. I always thought not(0)=1, so (1+1)=2=True. What's wrong in this logic?

Aakash aggarwal Over a year ago

think of it as not 0 + 1 I guess that will answer your question

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