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'2'<'1'== False #False
('2'<'1')== False #True
'2'<('1'== False) #error
code in python3 we know operator precedence in python3 https://docs.python.org/3/reference/expressions.html#operator-precedence
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This is because on the ASCII (American Standard Code For Information Interchange) CHART the letter "a" equates to 97 (in decimal values) while the letter "b" equates to 98 (in the decimal values).
Numbers. In Python, the integer 0 is always False , while every other number, including negative numbers, are True .
Python assigns boolean values to values of other types. For numerical types like integers and floating-points, zero values are false and non-zero values are true. For strings, empty strings are false and non-empty strings are true.
Answer: There 3 ways to check if true false in Python. Let’s see the syntax for an If statement using a boolean. Example code if true false example code. Simple python example code. Boolean values are the two constant objects False and True. It has the value False.
All other values are considered true — so objects of many types are always true. Operations and built-in functions that have a Boolean result always return 0 or False for false and 1 or True for true, unless otherwise stated. (Important exception: the Boolean operations or and and always return one of their operands.)
first ('1' == False) is evaluated which is False but now the operation is '2'<False which is an illegal operation in python It is because the default behaviour for equality comparison (== and !=) is based on the identity of the objects, and as object False and object '1' does not share the same identity thus the result will be False
On the other hand, True is treated as 1 in many cases (and False as 0) which you can see when you do stuff like: Boolean values are the two constant objects False and True. They are used to represent truth values (although other values can also be considered false or true).
'2'<'1'== False
it is evaluated as '2'<'1' and '1' == False according to the operator chaining {thanks @ymonad to provide this link}
which will evaluated to be False
('2'<'1')== False
as () have higher precedence so will be evaluated first. so the expression will be reduced to False == False which will evaluate to be True
'2'<('1'== False)
first ('1' == False) is evaluated which is False but now the operation is '2'<False which is an illegal operation in python
EDIT:
To answer a question raised by @snr in the comment section
The vital question is why '1'== False is valid while '2'< bool is not
It is because the default behaviour for equality comparison (== and !=) is based on the identity of the objects, and as object False and object '1' does not share the same identity thus the result will be False
whereas a default behaviour of other comparison (<, >, <=, and >=) is not provided thus an attempt to do so raises TypeError
you can find this in documentation link provided by OP (under the heading value-comparisions)
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