Question

Write a python program

write a function numDigits(num) which counts the digits in int num. Answer code is given in the Answer tab (and starting code), which converts num to a str, then uses the len() function to count and return the number of digits.

Note that this does NOT work correctly for num < 0. (Try it and see.) Fix this in two different ways, by creating new functions numDigits2(num) and numDigits3(num) that each return the correct number of digits for all values of num.

First try modifying numDigits(num) to create numDigits2(num). Do so by checking the sign of num and returning the correct value if num < 0.

Next, try a different approach, implementing your solution in numDigits3(num). Use a while loop and accumulator variable, repeatedly dividing num by 10 (use //) and incrementing the variable, until 0 is reached. Then return the value of the variable. But... there's a flaw in this approach for num==0 and others; make sure you fix them.

#
# Given Starter Code
#

def numDigits(num):
n_str = str(num)
return len(n_str)

def numDigits2(num):
return -1

def numDigits3(num):
# use a while loop here
return -1

def main():

number = int(input("Enter a integer: "))

print (number, 'has', numDigits(number), 'digits.')

print("%s has %d digits (different output)." % (number, numDigits(number))) # alternate output
print("%s has %d digits." % (number, numDigits2(number))) # alternate output
print("%s has %d digits." % (number, numDigits3(number))) # alternate output

# in the above, % is the old Python formatting operator:
#
# fmt-string % (v1,v2,...) results in each vN substituted into % slots within
# fmt-string, with slot types %d for int, %s for str, %f for float
#

main()

Answer 1

CODE :

#
# Given Starter Code
#
import math
def numDigits(num):
n_str = str(num)
return len(n_str)

def numDigits2(num):
n_str = str(num)
#If number is less than 0 return one less than the string length
if num<0:
return len(n_str)-1
else:
return len(n_str)

def numDigits3(num):
# use a while loop here
ans=0
#If number is 0 return 1
if num == 0 :
return 1
#If number is less than 0 , "//" gives floor of a number and gives wrong result
#Hence we use ceil function to using normal division
if num<0:
while num != 0:
num = num/10
num = math.ceil(num)
ans+=1
#If number is greater than 0 , then its as stated in the question
else:
while num != 0:
num = num//10
ans+=1
return ans

def main():

number = int(input("Enter a integer: "))
print (number, 'has', numDigits(number), 'digits.')
print("%s has %d digits (different output)." % (number, numDigits(number))) # alternate output
print("%s has %d digits." % (number, numDigits2(number))) # alternate output
print("%s has %d digits." % (number, numDigits3(number))) # alternate output
  
# in the above, % is the old Python formatting operator:
#
# fmt-string % (v1,v2,...) results in each vN substituted into % slots within
# fmt-string, with slot types %d for int, %s for str, %f for float
#

main()