I need a code in Python 3.3 to convert an integer into binary. This is my first try:
a = input(str("Please Enter a Number")
if a == float:
print (1)
else print(0)
b = a/2
while True:
if b == float:
print(1)
else print(0)
I don't know why I keep getting errors with the if a == float:.
And I know that the rest of the code is wrong too, but this : makes me crazy.
Your code has a lot of issues:
isinstance to see if an object is a float. I assume this is what you are trying to do with a == float. But, that doesn't make sense because, in Python 3.x., input always returns a string object. So, a is a string. However, if float is actually a variable, then you should change its name. Naming a variable float is a bad practice since it overrides the built-in.else.str in the first line is unnecessary (not an error, but I just thought I'd mention it).However, instead of fixing all this, I'm going to introduce you to the bin built-in:
>>> n = 127
>>> bin(n)
>>> # The "0b" at the start means "binary".
'0b1111111'
>>> # This gets rid of the "0b"
>>> bin(n)[2:]
'1111111'
>>>
It was built explicitly to do what you are trying to do.
Also, here are some references on Python you might enjoy:
a == float is a type test. He might actually have a variable named float.You can just use the bin function:
>>> bin(100)
'0b1100100'
Ignore the 0b infront of the string. You can always get the raw binary numbers using using bin(your_numer)[2:].
Also, you can get this using the format function:
>>> format(100, 'b')
'1100100'
If you need to print it in binary you can just do: print(bin(a))
This is what i made
while True:
print("FIND OUT WHAT BINARY THIS IS")
space = " "
num1 = int(input())
while num1 > 0:
if num1 % 2 == 0:
space = space + "0"
else:
space = space + "1"
num1 = int(num1 / 2)
else:
space = space[::-1]
print(space)
a == floatdo what I want?" Since you didn't ask a question, it is hard to know which question to answer. Please edit your post to include a specific question.