Dealing With Very Large Hex Numbers In Python

I need to convert an extremely large number (760402852596084587359490684321824034940816612213847025986535451828145781910762684416) to hexadecimal in python, but it seems to round

Solution 1:

Are you sure it's wrong? you can test the code below. The results are the same:

def make_hex(a):
    list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f']
    output = []
    while a>1:
        output.append(list[a%16])
        a = a//16output.reverse()   
    return'0x' + ''.join(output)

a = 760402852596084587359490684321824034940816612213847025986535451828145781910762684416print(make_hex(a))
print(hex(a))

Solution 2:

I tried your original number by converting it with a different function and then adding 1 by 1.

Answer: Python isn't truncating, the hex of your number just happens to end in 0s:

>>>num=760402852596084587359490684321824034940816612213847025986535451828145781910762684416>>>to_bytes(num)
'643437346d684000000000000000000000000000000000000000000000000000000000'
>>>to_bytes(num+1)
'643437346d684000000000000000000000000000000000000000000000000000000001'
>>>to_bytes(num+2)
'643437346d684000000000000000000000000000000000000000000000000000000002'

Here's my to_bytes function for reference:

def to_bytes(i, count=0, endian='big'):
    count = 1if i < 256elsemath.ceil(math.log(i + 1, 256))
    return i.to_bytes(count, endian).hex()