There are two methods:

• float_number = float(decimal_number)
• float_number = decimal_number * 1.0

First one using the built in function and second one without any function, but take care about the 1.0 not 1 as it should be float so the result will be float.

You can just do a cast

int val1 = 1;
float val2 = (float)val1;

or

decimal val3 = 3;
float val4 = (float)val3;

Python <2.7

"%.15g" % f

Or in Python 3.0:

format(f, ".15g")

Python 2.7+, 3.2+

Just pass the float to Decimal constructor directly, like this:

from decimal import Decimal
Decimal(f)

For the exponent $c$ that has indeed to be integer, you compute the next largest integer to the binary logarithm, $\log_2(1.5\cdot 10^8)=27.160..$ Then for the mantissa you need the integer part of $1.5\cdot 10^8\cdot 2^{20-28}=585937.5$. Now to get the digits you need to convert this decimal number to binary. One step there could be the hexadecimal form ${\tt 8F0D2}$.