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i want convert this number to binary so i can convert it to floating point later, the algorithm i know takes the part after the point of a number, for example if i put 9.32 it takes 0.32 and divides it by 0.5 on the function puntoflot() when its divided by 0.5 it takes the first number and puts it on z then takes the decimals left and keeps dividing it, you can figure it out there in the code.... idk if this algorithm is good any help would be apreciated

the valorbinario() function works awesomely with a big number like 9.32 BUT with a number like this 9.10938291*10**-31 it doesnt run valorbinario() at all because it says this error 

Traceback (most recent call last):
  File "<pyshell#8>", line 1, in <module>
    valorbinario(9.10938291*10**-31)
  File "C:/Users/Emmanuel/Desktop/asdfad.py", line 26, in valorbinario
    x=puntoflot(numeroderecha)
  File "C:/Users/Emmanuel/Desktop/asdfad.py", line 11, in puntoflot
    dere=int(x[2:])
ValueError: invalid literal for int() with base 10: '18765820000004e-32'

the puntoflot() recieves the part after the point for example 9.28 it takes 0.28 form valorbinario() valorbinario takes a float number and does alot of things, it substracts the decimals and puts it on the function puntoflot() in order to convert it to binary then converts the natural number in this case 9 to binary and then returns the whole number in this case '1001.010001111010111011001100'

so how can i do this?? if you can come up with a better algortihm to convert to binary i would be happy but if not, how can i fix this??

ok this is my code

you run valorbinario() first with a number the other function idk why i made it a function.... im just starting to learn code

def puntoflot(y):
    i=0
    z=""
    while i<=23:
        x=str(y/0.5)
        z=z+str(x[0])
        dere=int(x[2:])
        y=float("0."+str(dere))
        i+=1
    return z


def valorbinario(numero):
    numero=str(numero)
    pos=numero.index(".")
    numeroderecha=float("0."+str(numero[pos+1:]))
    numeroizquierda=numero[:pos]
    y=bin(int(numeroizquierda))
    x=puntoflot(numeroderecha)
return y[2:]+"."+str(x)
DestinyClouds
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  • "i want convert this number to binary so i can convert it to floating point later" - but it's already floating-point. You can scrap literally all of this code. – user2357112 May 16 '15 at 02:56
  • mmm no i want to convert this number to binary and then in binary i want to do IEEE754 but i want to do this manually i just need to know the binary number – DestinyClouds May 16 '15 at 03:00

1 Answers1

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For ints, bin(numero) works. For floats less than 1, you can try this:

bin(int(1/(numero)))

That will give you the inverse (1/x) of the number and turn it into an int, which you can then invert again later.

TigerhawkT3
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