Converting Decimal to Hexadecimal in python

Question

i am getting answer as 7D while i should be getting 7D0. I actually dont know how to solve it:
my test case at the beginning was 31 which answer was correct 1F but as i go towards higher number my code doesnt work.
using loops might be more efficient but i am not too comfortable with loop right now

please help !!!

def dec2hex(n):
x1 =0
counter = 0 
answer = ""
if n<=0:
    answer =answer + "0"
else:
    while (n >16):
        counter +=1
        n  =  n /16

        x = n

        if(x <16 ):
            x = int(n)
            break
        else: 
            continue
    if ((n-x) *16 <16 ):
        counter1 = 1
    else:
        counter1 = counter -1
    rem = (n-x) * (16**(counter1))

    if rem >16:
        while (n >16):
            rem = rem /16

            x1 = rem

            if(x1 <16 ):
                x1 = int(rem)
                break
            else: 
                continue
    if n < 10:
        answer =answer + str(int(x))
    if (rem  ==10 or x1 ==10):
        answer = answer + "A"
    if (rem  ==11 or x1 ==11):
        answer = answer + "B"
    if (rem ==12 or x1 ==12):
        answer = answer + "C"
    if (rem  ==13 or x1 ==13):
        answer = answer + "D"
    if (rem  ==14 or x1 ==14):
        answer = answer + "E"
    if (rem  ==15 or x1 ==15):
        answer = answer + "F"
    print(counter,rem,x1,n,counter,x)
return answer

dec2hex(2000)

If this isn't an exercise, you'd simply want return f"{n:X}" — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Feb 14, 2018 at 0:33
Note, you are using loops. The simplest way would be to accumulate the hexidecimal digits (strings) in a list. Start by successively taking the remainder and appending the corresponding hexi-digit (you could use a list for this). Then, reverse the final result and ''.join itno a string. You really only need a single while-loop — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Feb 14, 2018 at 0:40
Note, to get the remainder simply use the modulus operator, i.e. %, no need for this: rem = (n-x) * (16**(counter1)) you don't need a counter. So, simply put, you could do, n, rem = n // 16, n % 16, or use the helper function divmod, so n, rem = divmod(n, 16) — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Feb 14, 2018 at 0:41
I am sorry i just noticed the mistake in the heading, i need to convert decimal to Hexadecimal no the other way around @juanpa.arrivillaga — shiv
– shiv, Commented Feb 14, 2018 at 0:54

Pyroseza · Accepted Answer · 2019-08-15 21:21:10Z

1

The most elegant answer I can think of is to format a string literal and it will convert the value for you

3.6+ (f-strings)

>>> d = 2000
>>> print(f'{d:X}')
7D0

pre-3.6 (string format function)

>>> d = 2000
>>> print('{:X}'.format(d))
7D0

https://docs.python.org/3/reference/lexical_analysis.html#f-strings

https://docs.python.org/3/library/string.html#formatspec

answered Aug 15, 2019 at 21:21

Pyroseza

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Comments

AndroX · Accepted Answer · 2018-05-16 07:14:46Z

I would like to propose this approach to the problem. A while loop is used to cycle the main division and the reminder calculus A for loop is used to invert the final answer string

The while loop condition exclude the case in which 0 is entered becasue the division remainder will be always zero and the loop will be infinite. In this case the answer will be forced to 0 by the first if.

def dec2hex(hexnum):
hexstring = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"]
counter = 0
remainder = []
answer = ""
if hexnum > 0:
    while hexnum > 0:
        remainder.append(hexnum % 16)
        hexnum = hexnum // 16
        counter = counter + 1
    for reverse in remainder[::-1]:
        answer = answer + hexstring[reverse]
else:
    answer = "0"
return answer

print(dec2hex(2000))

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