[python] working with negative numbers in python

I am a student in a concepts of programming class. The lab is run by a TA and today in lab he gave us a real simple little program to build. It was one where it would multiply by addition. Anyway, he had us use absolute to avoid breaking the prog with negatives. I whipped it up real quick and then argued with him for 10 minutes that it was bad math. It was, 4 * -5 does not equal 20, it equals -20. He said that he really dosen't care about that and that it would be too hard to make the prog handle the negatives anyway. So my question is how do I go about this.

here is the prog I turned in:

#get user input of numbers as variables

numa, numb = input("please give 2 numbers to multiply seperated with a comma:")

#standing variables
total = 0
count = 0

#output the total
while (count< abs(numb)):
    total = total + numa
    count = count + 1

#testing statements
if (numa, numb <= 0):
    print abs(total)
else:
    print total

I want to do it without absolutes, but every time I input negative numbers I get a big fat goosegg. I know there is some simple way to do it, I just can't find it.

The abs() in the while condition is needed, since, well, it controls the number of iterations (how would you define a negative number of iterations?). You can correct it by inverting the sign of the result if numb is negative.

So this is the modified version of your code. Note I replaced the while loop with a cleaner for loop.

#get user input of numbers as variables
numa, numb = input("please give 2 numbers to multiply seperated with a comma:")

#standing variables
total = 0

#output the total
for count in range(abs(numb)):
    total += numa

if numb < 0:
    total = -total

print total

How about something like that? (Uses no abs() nor mulitiplication)
Notes:

• the abs() function is only used for the optimization trick. This snippet can either be removed or recoded.
• the logic is less efficient since we're testing the sign of a and b with each iteration (price to pay to avoid both abs() and multiplication operator)

def multiply_by_addition(a, b):
""" School exercise: multiplies integers a and b, by successive additions.
"""
   if abs(a) > abs(b):
      a, b = b, a     # optimize by reducing number of iterations
   total = 0
   while a != 0:
      if a > 0:
         a -= 1
         total += b
      else:
         a += 1
         total -= b
   return total

multiply_by_addition(2,3)
6
multiply_by_addition(4,3)
12
multiply_by_addition(-4,3)
-12
multiply_by_addition(4,-3)
-12
multiply_by_addition(-4,-3)
12

Thanks everyone, you all helped me learn a lot. This is what I came up with using some of your suggestions

#this is apparently a better way of getting multiple inputs at the same time than the 
#way I was doing it
text = raw_input("please give 2 numbers to multiply separated with a comma:")
split_text = text.split(',')
numa = int(split_text[0])
numb = int(split_text[1])

#standing variables
total = 0

if numb > 0:
    repeat = numb
else:
    repeat = -numb

#for loops work better than while loops and are cheaper
#output the total
for count in range(repeat):
    total += numa


#check to make sure the output is accurate
if numb < 0:
    total = -total


print total

Thanks for all the help everyone.

Too hard? Your TA is... well, the phrase would probably get me banned. Anyways, check to see if numb is negative. If it is then multiply numa by -1 and do numb = abs(numb). Then do the loop.

import time

print ('Two Digit Multiplication Calculator')
print ('===================================')
print ()
print ('Give me two numbers.')

x = int ( input (':'))

y = int ( input (':'))

z = 0

print ()


while x > 0:
    print (':',z)
    x = x - 1
    z = y + z
    time.sleep (.2)
    if x == 0:
        print ('Final answer: ',z)

while x < 0:
    print (':',-(z))
    x = x + 1
    z = y + z
    time.sleep (.2)
    if x == 0:
        print ('Final answer: ',-(z))

print ()  

Try this on your TA:

# Simulate multiplying two N-bit two's-complement numbers
# into a 2N-bit accumulator
# Use shift-add so that it's O(base_2_log(N)) not O(N)

for numa, numb in ((3, 5), (-3, 5), (3, -5), (-3, -5), (-127, -127)):
    print numa, numb,
    accum = 0
    negate = False
    if numa < 0:
        negate = True
        numa = -numa
    while numa:
        if numa & 1:
            accum += numb
        numa >>= 1
        numb <<= 1
    if negate:
        accum = -accum
    print accum

output:

3 5 15
-3 5 -15
3 -5 -15
-3 -5 15
-127 -127 16129