Length of hexadecimal number
Question:
How can we get the length of a hexadecimal number in the Python language?
I tried using this code but even this is showing some error.
i = 0
def hex_len(a):
if a > 0x0:
# i = 0
i = i + 1
a = a/16
return i
b = 0x346
print(hex_len(b))
Here I just used 346 as the hexadecimal number, but my actual numbers are very big to be counted manually.
Answers:
Use the function hex
:
>>> b = 0x346
>>> hex(b)
'0x346'
>>> len(hex(b))-2
3
or using string formatting:
>>> len("{:x}".format(b))
3
As Ashwini wrote, the hex
function does the hard work for you:
hex(x)
Convert an integer number (of any size) to a hexadecimal string. The result is a valid Python expression.
While using the string representation as intermediate result has some merits in simplicity it’s somewhat wasted time and memory. I’d prefer a mathematical solution (returning the pure number of digits without any 0x-prefix):
from math import ceil, log
def numberLength(n, base=16):
return ceil(log(n+1)/log(base))
The +1 adjustment takes care of the fact, that for an exact power of your number base you need a leading “1”.
How can we get the length of a hexadecimal number in the Python language?
I tried using this code but even this is showing some error.
i = 0
def hex_len(a):
if a > 0x0:
# i = 0
i = i + 1
a = a/16
return i
b = 0x346
print(hex_len(b))
Here I just used 346 as the hexadecimal number, but my actual numbers are very big to be counted manually.
Use the function hex
:
>>> b = 0x346
>>> hex(b)
'0x346'
>>> len(hex(b))-2
3
or using string formatting:
>>> len("{:x}".format(b))
3
As Ashwini wrote, the hex
function does the hard work for you:
hex(x)
Convert an integer number (of any size) to a hexadecimal string. The result is a valid Python expression.
While using the string representation as intermediate result has some merits in simplicity it’s somewhat wasted time and memory. I’d prefer a mathematical solution (returning the pure number of digits without any 0x-prefix):
from math import ceil, log
def numberLength(n, base=16):
return ceil(log(n+1)/log(base))
The +1 adjustment takes care of the fact, that for an exact power of your number base you need a leading “1”.