CODEWITHSUNDEEP
pythoncomparehex
Bill
Bill

Reputation: 53

Python compare char to hex

>>> var = 'g'  
>>> print hex(ord(var))  
0x67  
>>> print hex(ord(var)) == 0x67  
False

Why isn't this true in python 2.7?
What would be the best way to compare 'g' to the hex value 0x67?

Upvotes: 2

Views: 3638

Answers (4)

Ahasanul Haque
Ahasanul Haque

Reputation: 11134

First see the type of hex(ord(var)):

>>> print type(hex(ord(var)))
<type 'str'>

Then see the type of 0x67

>>> type(0x67)
<type 'int'>

You are comparing a str with an int. So, you should do:

hex(ord(var)) == '0x67'

Upvotes: 0

Lol4t0
Lol4t0

Reputation: 12547

According to the documentation

hex(x)

Convert an integer number (of any size) to a lowercase hexadecimal string prefixed with 0x

So hex(ord(var)) == '0x67'

It just print that removes quotes.

See

>>> var = 'g'
>>> hex(ord(var))
'0x67'
>>> hex(0x67)
'0x67'
>>> hex(ord(var)) == hex(0x67)
True

And of course ord(g) == 0x67 because numbers are equal despite of representation, i.e 0x67 and 103 and 0147 are all the same number internally

Upvotes: 0

idjaw
idjaw

Reputation: 26580

You can simply take the ord and compare it to 0x67

>>> ord('g') == 0x67
True

If you do this: 

>>> 0x67
103

You are still getting the ascii code for that character.

Furthermore, based on your explicit example, if you are trying to cast that to an int to actually compare to 0x67, then you need to do it in base-16: 

>>> int(hex(ord('g')), 16) == 0x67
True

False case:

>>> int(hex(ord('d')), 16) == 0x67
False

Upvotes: 0

Matteo Italia
Matteo Italia

Reputation: 126867

hex returns a string, which you are comparing to a number. Either do 

ord(var) == 0x67

or

hex(ord(var)) == "0x67"

(the first one is less error-prone, as it's case insensitive)

Upvotes: 1

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