How to convert an int to a hex string?

I want to take an integer (that will be <= 255), to a hex string representation

e.g.: I want to pass in 65 and get out '\x41', or 255 and get '\xff'.

I've tried doing this with the struct.pack('c',65), but that chokes on anything above 9 since it wants to take in a single character string.

14 Answers

You are looking for the chr function.

You seem to be mixing decimal representations of integers and hex representations of integers, so it's not entirely clear what you need. Based on the description you gave, I think one of these snippets shows what you want.

>>> chr(0x65) == '\x65' True >>> hex(65) '0x41' >>> chr(65) == '\x41' True 

Note that this is quite different from a string containing an integer as hex. If that is what you want, use the hex builtin.

This will convert an integer to a 2 digit hex string with the 0x prefix:

strHex = "0x%0.2X" % integerVariable 

What about hex()?

hex(255) # 0xff 

If you really want to have \ in front you can do:

print '\\' + hex(255)[1:] 

Let me add this one, because sometimes you just want the single digit representation

( x can be lower, 'x', or uppercase, 'X', the choice determines if the output letters are upper or lower.):

'{:x}'.format(15) > f 

And now with the new f'' format strings you can do:

f'{15:x}' > f 

To add 0 padding you can use 0>n:

f'{2034:0>4X}' > 07F2 

NOTE: the initial 'f' in f'{15:x}' is to signify a format string

Try:

"0x%x" % 255 # => 0xff 

or

"0x%X" % 255 # => 0xFF 

Python Documentation says: "keep this under Your pillow: "

If you want to pack a struct with a value <255 (one byte unsigned, uint8_t) and end up with a string of one character, you're probably looking for the format B instead of c. C converts a character to a string (not too useful by itself) while B converts an integer.

struct.pack('B', 65) 

(And yes, 65 is \x41, not \x65.)

The struct class will also conveniently handle endianness for communication or other uses.

With format(), as per format-examples, we can do:

>>> # format also supports binary numbers >>> "int: {0:d}; hex: {0:x}; oct: {0:o}; bin: {0:b}".format(42) 'int: 42; hex: 2a; oct: 52; bin: 101010' >>> # with 0x, 0o, or 0b as prefix: >>> "int: {0:d}; hex: {0:#x}; oct: {0:#o}; bin: {0:#b}".format(42) 'int: 42; hex: 0x2a; oct: 0o52; bin: 0b101010' 

Note that for large values, hex() still works (some other answers don't):

x = hex(349593196107334030177678842158399357) print(x) 

Python 2: 0x4354467b746f6f5f736d616c6c3f7dL
Python 3: 0x4354467b746f6f5f736d616c6c3f7d

For a decrypted RSA message, one could do the following:

import binascii hexadecimals = hex(349593196107334030177678842158399357) print(binascii.unhexlify(hexadecimals[2:-1])) # python 2 print(binascii.unhexlify(hexadecimals[2:])) # python 3 

This worked best for me

"0x%02X" % 5 # => 0x05 "0x%02X" % 17 # => 0x11 

Change the (2) if you want a number with a bigger width (2 is for 2 hex printned chars) so 3 will give you the following

"0x%03X" % 5 # => 0x005 "0x%03X" % 17 # => 0x011 

(int_variable).to_bytes(bytes_length, byteorder='big'|'little').hex() 

For example:

>>> (434).to_bytes(4, byteorder='big').hex() '000001b2' >>> (434).to_bytes(4, byteorder='little').hex() 'b2010000' 

I wanted a random integer converted into a six-digit hex string with a # at the beginning. To get this I used

"#%6x" % random.randint(0xFFFFFF) 

Also you can convert any number in any base to hex. Use this one line code here it's easy and simple to use:

hex(int(n,x)).replace("0x","")

You have a string n that is your number and x the base of that number. First, change it to integer and then to hex but hex has 0x at the first of it so with replace we remove it.

For Python >= 3.6, use f-string formatting:

>>> x = 114514 >>> f'{x:0x}' '1bf52' >>> f'{x:#x}' '0x1bf52' 

As an alternative representation you could use

[in] '%s' % hex(15) [out]'0xf'