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 
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".
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.
This is the program to count the num of digits in hex number using recursion
def count_hex(n): ''' returns the num of digits in hex''' if n == 0: return 0 n >>= 4 return 1 + count_hex(n) print(count_hex(0xFFFF))
4
You could also do this without converting to a string using the bit_length method:
def hexLen(N): return (N.bit_length()+3)//4 Note that his returns a length of zero when N is 0, your function would return None but, in reality, you still need one hex digit to represent zero. (hint: add max(1,...) to cover that edge case)
0x346to be?