How to work with hex() in python?

Question

I've problem with hex() in python 2.7 I want to convert '0777' to hex with user input. but it have problem with using integer with user input.

In [1]: hex(0777)
Out[1]: '0x1ff'

In [2]: hex(777)
Out[2]: '0x309'

In [3]: z = raw_input('enter:')
enter:0777

In [4]: z
Out[4]: '0777'

In [5]: hex(z)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-3682d79209b9> in <module>()
----> 1 hex(z)


TypeError: hex() argument can't be converted to hex

In [6]: hex(int(z))
Out[6]: '0x309'

In [7]:

I need 0x1ff but its showing me 0x309, how i can fix it ?

so you specifically only want octal input? int('8',8) -> error — Padraic Cunningham
– Padraic Cunningham, Commented May 17, 2015 at 11:23

vaultah · Accepted Answer · 2015-05-17 11:32:31Z

4

The base argument of the int class defaults to 10

int(x, base=10) -> integer

leading zeroes will be stripped. See this example:

In [1]: int('0777')
Out[1]: 777

Specify base 8 explicitly, then the hex function will give you the desired result:

In [2]: hex(int('0777', 8))
Out[2]: '0x1ff'

edited May 17, 2015 at 11:32

answered May 17, 2015 at 11:14

vaultah

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Comments

Assem · Accepted Answer · 2015-05-17 11:20:54Z

0

You can use input() instead of raw_input() to eval input and read octal values.

In [3]: z = input('enter:')
enter:0777

In [4]: z
Out[4]: 511

In [5]: hex(z)'
Out[5]: '0x1ff'

answered May 17, 2015 at 11:20

Assem

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3 Comments

Padraic Cunningham Over a year ago

Don't useinput ever

Assem Over a year ago

if his input is trust-able, why not?

Padraic Cunningham Over a year ago

how do you make user input trust-able?

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