Prerequisite : IEEE Standard 754 Floating Point Numbers
Given a floating point number, the task is to find the hexadecimal representation for the number by IEEE 754 standard.
The IEEE Standard for Floating-Point Arithmetic (IEEE 754) is a technical standard for floating-point computation which was established in 1985 by the Institute of Electrical and Electronics Engineers (IEEE). The standard addressed many problems found in the diverse floating point implementations that made them difficult to use reliably and reduced their portability. IEEE Standard 754 floating point is the most common representation today for real numbers on computers, including Intel-based PC’s, Macs, and most Unix platforms.
Examples :
Input : -6744.90 Output : C5D2C733 Input : -263.3 Output : C383A666
Approach :
- Check whether the number is positive or negative. Save the sign as 0 for positive and 1 for negative, and then convert the number into positive if it is negative.
- Convert the floating point number to binary.
- Separate the decimal part and the whole number part.
- Calculate the exponent(E) and convert it to binary.
- Find the mantissa.
- Concatenate the sign of mantissa, exponent and the mantissa.
- Convert it into hexadecimal.
Let’s write a Python program to represent a floating number as hexadecimal by IEEE 754 standard.
Python3
# Function for converting decimal to binarydef float_bin(my_number, places = 3): my_whole, my_dec = str(my_number).split(".") my_whole = int(my_whole) res = (str(bin(my_whole))+".").replace('0b','') for x in range(places): my_dec = str('0.')+str(my_dec) temp = '%1.20f' %(float(my_dec)*2) my_whole, my_dec = temp.split(".") res += my_whole return resdef IEEE754(n) : # identifying whether the number # is positive or negative sign = 0 if n < 0 : sign = 1 n = n * (-1) p = 30 # convert float to binary dec = float_bin (n, places = p) dotPlace = dec.find('.') onePlace = dec.find('1') # finding the mantissa if onePlace > dotPlace: dec = dec.replace(".","") onePlace -= 1 dotPlace -= 1 elif onePlace < dotPlace: dec = dec.replace(".","") dotPlace -= 1 mantissa = dec[onePlace+1:] # calculating the exponent(E) exponent = dotPlace - onePlace exponent_bits = exponent + 127 # converting the exponent from # decimal to binary exponent_bits = bin(exponent_bits).replace("0b",'') mantissa = mantissa[0:23] # the IEEE754 notation in binary final = str(sign) + exponent_bits.zfill(8) + mantissa # convert the binary to hexadecimal hstr = '0x%0*X' %((len(final) + 3) // 4, int(final, 2)) return (hstr, final)# Driver Codeif __name__ == "__main__" : print (IEEE754(263.3)) print (IEEE754(-263.3)) |
4383A666 C383A666
Time complexity: The time complexity of this function is O(p) where p is the number of places of the decimal value passed to the function. This is because the function uses a for loop that runs for p iterations to convert the decimal value to binary.
Space complexity: The space complexity of this function is O(1), as the function uses only a few variables that are independent of the input size. The function does not use any data structures with size dependent on the input.
