# Python binary calculator

My task was to build a calculator in Python supporting binary addition etc.

To begin I define a couple of custom exceptions :

class MaximumValueException(Exception):
pass

class MinimumValueException(Exception):
pass

I then define the Binary class which does most of the work for me (I know that the division is not totally working yet but everything else is) :

class Binary:
def __init__(self, application, integer: int = None):
self.app = application
self.binary_value = self.convert_int_to_binary(integer) if integer is not None else None

def from_binary(self, binary: str):  # init from binary as opposed to an integer
self.binary_value = self.allocate_bits(binary[::-1])[::-1]
return self

def __str__(self):
return self.binary_value

def __len__(self):
return len(self.binary_value)

first = list(self.binary_value[::-1])  # flips both around to work with left to right rather than opposite
second = list(other.binary_value[::-1])

carry = False

for i in range(0, len(first)):
if carry:

carry = False
else:
carry = True

def __sub__(self, other):
return self + (- other)

def __mul__(self, other):
other_value_reversed = list(other.binary_value[::-1])
total = Binary(self.app, 0)
for i in range(1, len(other_value_reversed) + 1):
binary_val = int(self.binary_value) * int(other_value_reversed[i - 1]) * int(10 ** (i - 1))
total = total + Binary(self.app).from_binary(str(binary_val))

def __ge__(self, other):
return self.denary() >= other.denary()  # I know this isn't ideal but will come back around to it.

def __le__(self, other):
return self.denary() <= other.denary()

def __truediv__(self, divisor):
dividend = self.binary_value
cur_div = ""

for i in range(0, len(dividend)):
cur_div = cur_div + dividend[i]
if divisor <= Binary(self.app).from_binary(cur_div):
cur_div = (Binary(self.app).from_binary(cur_div) - take_away).binary_value

else:

return self

def __neg__(self):
return self.change_state()

def denary(self):
binary = list(self.binary_value)  # flips string
value = 0
if binary[0] == "1":
binary = list(self.change_state().binary_value[::-1])
for i in range(0, len(binary) - 1):
value += ((2 ** i) * int(binary[i]))
return -int(value)
elif binary[0] == "0":
binary = list(self.binary_value[::-1])
for i in range(0, len(binary) - 1):
value += ((2 ** i) * int(binary[i]))
return int(value)

def convert_int_to_binary(self, number):
neg = False
if number < 0:
neg = True
binary = self.positive_binary_convert(abs(number))
if neg:
binary = binary.change_state()
return binary.binary_value

def positive_binary_convert(self, integer):
before = ""
finished = False
while not finished:
divided = int(int(integer) / 2)
remainder = int(integer) % 2
before = before + str(remainder)
integer = divided
if divided == 0:
finished = True
before = self.allocate_bits(before)
binary = before[::-1]  # Flips the string
return Binary(self.app).from_binary(binary)

def change_state(self, binary=None):  # Change from negative to positive
if binary is not None:
flipped = Binary(self.app).from_binary(binary).flip_bits()
else:
flipped = self.flip_bits()

return flipped + Binary(self.app, 1)

def allocate_bits(self, value):
length = len(value)
if length < self.app.bits:
rem = self.app.bits - length
else:
rem = length % self.app.bits
for i in range(0, rem):
value = value + "0"
return value

def flip_bits(self):
binary_list = list(self.binary_value)

for i in range(0, len(self.binary_value)):
binary_list[i] = "1" if binary_list[i] == "0" else "0"
binary = "".join(binary_list)
return Binary(self.app).from_binary(binary)

Finally I define a Application class:

class Application:
def __init__(self, bits=8):
self.bits = bits
self.max_bound = (2 ** (self.bits - 1)) - 1
self.min_bound = self.max_bound * -1

def welcome_message(self):
print("Welcome to the Binary Calculator!")
print("---------------------------------")
print("This Calculator is currently set to " + str(self.bits) + " bits.")
print("The Maximum Bound of the calculator is " + str(self.max_bound))
print("The Minimum Bound of the calculator is " + str(self.min_bound))

def input_number(self, message):
while True:
try:
user_input = int(input(message))
if user_input > self.max_bound:
raise MaximumValueException()
if user_input < self.min_bound:
raise MinimumValueException()

except ValueError:
print("Not an integer! Try again.")
continue
except MaximumValueException:
print('Value above maximum bound of ' + str(self.max_bound))
except MinimumValueException:
print('Value below minimum bound of ' + str(self.min_bound))

else:
return user_input

@staticmethod
def input_sum_type():
return str(input("What do you want to do (add, take, divide, multiply)? "))

def input_integer(self, message):
denary = self.input_number(message)
binary = Binary(self, denary)
print(binary)
return binary

@staticmethod
def output_spacer():
print("--------")

@staticmethod
def calculate(sum_type, first_binary, second_binary):
return first_binary + second_binary
elif sum_type in ['take', '-']:
return first_binary - second_binary
elif sum_type in ['multiply', 'times', 'x', '*']:
return first_binary * second_binary
else:
return None

def output_calculated(self, calculated):
print(calculated)
self.output_spacer()
print(calculated.denary())

def retry(self):
try_again = str(input("Do you want to try again? "))
if try_again.lower() in ["yes", "y", "yea"]:
self.main()
else:
print("Thanks.")

def test(self):
print("Passed" if (Binary(self, 5) + Binary(self, 10)).denary() == 15 else "Failed")
print("Passed" if (Binary(self, 125) + Binary(self, 125)).denary() == 250 else "Failed")
print("Passed" if (Binary(self, 5) - Binary(self, 10)).denary() == -5 else "Failed")
print("Passed" if (Binary(self, 255) - Binary(self, 120)).denary() == 135 else "Failed")
print("Passed" if (Binary(self, 4) * Binary(self, 5)).denary() == 20 else "Failed")
print("Passed" if (Binary(self, 125) * Binary(self, 1000)).denary() == 125000 else "Failed")

def start(self):
self.welcome_message()
self.main()

def main(self):
sum_type = self.input_sum_type()
first_binary = self.input_integer("Enter your first Integer: ")
second_binary = self.input_integer("Enter your Second Integer: ")
self.output_spacer()
calculated = self.calculate(sum_type, first_binary, second_binary)
if calculated is not None:
self.output_calculated(calculated)
self.retry()

I then start the application:

Application(64).start()

General design comment

The fact that a binary number needs to access the application using it looks like a huge design issue.

A quick fix for this is to provide the desired numbers of bits instead - I am not sure how relevant this suggestion is but it was enough to get me started. Now, we have a Binary class with no dependencies and some tests for it:

def test(bits = 64):
test_equal(Binary(bits, 5) + Binary(bits, 10), 15)
test_equal(Binary(bits, 125) + Binary(bits, 125), 250)
test_equal(Binary(bits, 5) - Binary(bits, 10), -5)
test_equal(Binary(bits, 255) - Binary(bits, 120), 135)
test_equal(Binary(bits, 4) * Binary(bits, 5), 20)
test_equal(Binary(bits, 125) * Binary(bits, 1000), 125000)

The __add__ method is interesting because it gives a good chance to see many small techniques to improve Python code.

• In Place operators: for most operations (like +) there is a in-place version so that you can write a += b instead of a = a + b.

• In Python, Booleans True and False are actually integers that evaluate to 1 and 0. Instead of: added_val = e; if carry: added_val += 1, you could write added_val = e + carry.

• Look like a native: I highly recommend the article/talk from Ned Batchelder I've linked to. Basically, pretty much everytime you write are using indices and/or range(len(...)) in Python, there is a better way to do it using various Python tools.

In our case, we are using indices to loop over 2 iterables in parallel. This could be achieved with zip.

So far, we have:

first = list(self.binary_value[::-1])  # flips both around to work with left to right rather than opposite
second = list(other.binary_value[::-1])
assert len(first) == len(second)
carry = False

for i, j in zip(first, second):
added_val = int(i) + int(j) + carry

carry = False
else:
carry = True

Then, we could re-think the operations we are performing. For a start, we could reduce duplication by performing final_added += str(XXX) in a single place. We'd write:

carry = False
else:
carry = True

Because we are handling a number with is in (0, 1, 2, 3) at this stage (a sum of 3 bits), we are actually doing the equivalent of integer divisions by 2 by handling the different case. This can be written:

assert added_val in (0, 1, 2, 3)

Even better, Python has a shortcut to get both the quotient and the remainder of a division: divmod.

• We can get rid of useless operations: the conversions to list are not required: as far as we are concerned, we just need to iterate over things: it works if they are strings.

• We can also get rid of the [::-1] trick and just use reversed instead.

• Also, we don't even need the additional variables first and second anymore.

At this stage, we have

carry = False

for i, j in zip(reversed(self.binary_value), reversed(other.binary_value)):
carry, bit = divmod(int(i) + int(j) + carry, 2)
# TODO: Handle inputs of different length or carry overflow ?

It looks like it can hardly get any better, doesn't it ? Well it could: PEP 8, the Style Guide for Python recommends:

do not rely on CPython's efficient implementation of in-place string concatenation for statements in the form a += b or a = a + b. This optimization is fragile even in CPython (it only works for some types) and isn't present at all in implementations that don't use refcounting. In performance sensitive parts of the library, the ''.join() form should be used instead. This will ensure that concatenation occurs in linear time across various implementations.

Improving def __mul__(self, other)

Now that we've learnt a few Python tricks and best practises, we can have a look at __mul__ and re-apply most of them. Here again, we can get rid of range(len(XXX)), use in-place operator, use reversed, remove variables:

def __mul__(self, other):
total = Binary(self.bits, 0)
for i, val in enumerate(reversed(other.binary_value)):
total += Binary(self.bits).from_binary(str(int(self.binary_value) * int(val) * int(10 ** i)))

Now, something which may not be so interesting in this case but is always good to know is that the sum builtin can be use to rewrite such a expression in a more concise and more efficient way. In our case, it leads to an expression getting a bit too complicated:

def __mul__(self, other):
return sum(
(Binary(self.bits).from_binary(str(int(self.binary_value) * int(val) * int(10 ** i)))
for i, val in enumerate(reversed(other.binary_value))),
Binary(self.bits, 0))

Improving def denary(self)

Removing the duplicated logic (and introducing new variables), we could write:

def denary(self):
first_bit = self.binary_value[0]
if first_bit == "1":
sign = -1
newval = self.change_state()
elif first_bit == "0":
sign = 1
newval = self
value = 0
binary = newval.binary_value[::-1]
for i in range(0, len(binary) - 1):
value += ((2 ** i) * int(binary[i]))
return int(value) * sign

Now, it seems like the conversion int(value) is not required. Also, we could reuse our friend sum:

return sign * sum(((2 ** i) * int(binary[i])) for i in range(0, len(binary) - 1))

Now, we also can try to get rid of the range(len(XXX)) logic but before to do so, we have to understand what is going on: this time we do not handle all elements from binary, we stop before the last one. This is quite unexpected. Actually, it is easier to understand if we look at the whole method: we have handled the first character and we want to handle the remaining ones. The special character is not the last one, it is the first.

Keeping that mind, we can re-use the techniques already described: enumerate, reversed, etc.

def denary(self):
first_bit = self.binary_value[0]
remaining = self.binary_value[1:]
sign = 1
if first_bit == "1":
sign = -1
remaining = self.change_state().binary_value[1:]
return sign * sum(((2 ** i) * int(val)) for i, val in enumerate(reversed(remaining)))

Improving def allocate_bits(self, value):

We could use value += "0" but that would be still using the "in-place string concatenation" which is best to avoid when it is possible. In our case, we have a shorter and more efficient option: the mult operation on strings:

return value + "0" * rem

Improving def flip_bits(self)

Here again, we have the range(len(XXX)) pattern but we'll handle it slightly differently. Here, it looks like we are trying to update a list as we iterate. What we are actually trying to achieve is to build a new list based on something we can iterate on. We could use for bit in self.binary_value and list.append but it is a good chance to show off list comprehension:

binary_list = []
for bit in self.binary_value:
binary_list.append("1" if bit == "0" else "0" )

can also be written:

binary_list = ["1" if bit == "0" else "0" for bit in self.binary_value]

We get

def flip_bits(self):
binary = "".join(["1" if bit == "0" else "0" for bit in self.binary_value])
return Binary(self.bits).from_binary(binary)

Improving def positive_binary_convert(self, integer)

We can remove the finished variable: we could just break out of the loop for the time being.

We can remove the useless conversions int(integer).

We could once more use divmod.

We could use in-place operator.

At this stage, we have:

def positive_binary_convert(self, integer):
before = ""
while True:
divided, remainder = divmod(integer, 2)
before += str(remainder)
integer = divided
if divided == 0:
break

Also, the divided variable is not useful. We could use integer directly.

Improving def convert_int_to_binary(self, number)

You could use directly:

neg = number < 0

Also, you don't even need that variable at all:

def convert_int_to_binary(self, number):
binary = self.positive_binary_convert(abs(number))
if number < 0:
binary = binary.change_state()
return binary.binary_value

General comment

There is something I find particularly fishy in the from_binary method. From the name, it looks like it should be a class method (used as an alternative constuctor) but it is actually a normal method.

I've tried to re-write the code with a simple constructor and 2 alternative constructors (from int and from binary) and got the results below but I am still not fully convinced:

# https://codereview.stackexchange.com/questions/214533/python-binary-calculator

class Binary:
def __init__(self, bits):
self.bits = bits
self.binary_value = None

@classmethod
def from_int(cls, bits, integer: int):
self = cls(bits)
self.binary_value = self.convert_int_to_binary(integer)
return self

@classmethod
def from_binary(cls, bits, binary: str):
self = cls(bits)
self.binary_value = self.allocate_bits(binary[::-1])[::-1]
return self

def __str__(self):
return self.binary_value

def __len__(self):
return len(self.binary_value)

carry = False

for i, j in zip(reversed(self.binary_value), reversed(other.binary_value)):
carry, bit = divmod(int(i) + int(j) + carry, 2)
# TODO: Handle inputs of different length or carry overflow ?

def __sub__(self, other):
return self + (- other)

def __mul__(self, other):
return sum(
(Binary.from_binary(self.bits, str(int(self.binary_value) * int(val) * int(10 ** i)))
for i, val in enumerate(reversed(other.binary_value))),
Binary.from_int(self.bits, 0))

def __ge__(self, other):
return self.denary() >= other.denary()  # I know this isn't ideal but will come back around to it.

def __le__(self, other):
return self.denary() <= other.denary()

def __truediv__(self, divisor):
dividend = self.binary_value
cur_div = ""
for i in range(0, len(dividend)):
cur_div = cur_div + dividend[i]
if divisor <= Binary.from_binary(self.bits, cur_div):
take_away = divisor * Binary.from_binary(self.bits, answer)
cur_div = (Binary.from_binary(self.bits, cur_div) - take_away).binary_value

else:

return self

def __neg__(self):
return self.change_state()

def denary(self):
first_bit = self.binary_value[0]
remaining = self.binary_value[1:]
sign = 1
if first_bit == "1":
sign = -1
remaining = self.change_state().binary_value[1:]
return sign * sum(((2 ** i) * int(val)) for i, val in enumerate(reversed(remaining)))

def convert_int_to_binary(self, number):
binary = self.positive_binary_convert(abs(number))
if number < 0:
binary = binary.change_state()
return binary.binary_value

def positive_binary_convert(self, integer):
before = ""
while True:
integer, remainder = divmod(integer, 2)
before += str(remainder)
if integer == 0:
break
binary = self.allocate_bits(before)[::-1]  # Flips the string
return Binary.from_binary(self.bits, binary)

def change_state(self, binary=None):  # Change from negative to positive
if binary is not None:
flipped = Binary.from_binary(self.bits, binary).flip_bits()
else:
flipped = self.flip_bits()

return flipped + Binary.from_int(self.bits, 1)

def allocate_bits(self, value):
length = len(value)
if length < self.bits:
rem = self.bits - length
else:
rem = length % self.bits
return value + "0" * rem

def flip_bits(self):
binary = "".join(["1" if bit == "0" else "0" for bit in self.binary_value])
return Binary.from_binary(self.bits, binary)

def test_equal(binary_val, expected_res):
res = binary_val.denary()
if res == expected_res:
print("Passed")
else:
print("Failed", res, expected_res)

def test(bits = 64):
test_equal(Binary.from_int(bits, 5) + Binary.from_int(bits, 10), 15)
test_equal(Binary.from_int(bits, 125) + Binary.from_int(bits, 125), 250)
test_equal(Binary.from_int(bits, 5) - Binary.from_int(bits, 10), -5)
test_equal(Binary.from_int(bits, 255) - Binary.from_int(bits, 120), 135)
test_equal(Binary.from_int(bits, 4) * Binary.from_int(bits, 5), 20)
test_equal(Binary.from_int(bits, 125) * Binary.from_int(bits, 1000), 125000)

test()

• In denary you can write first_bit, *remaining = self.binary_value by using advanced tuple unpacking. Commented Mar 2, 2019 at 10:23