cs2010 Notes: 09-09 Base Conversions

Review: Digit places

Base 10: 1’s, 10’s, 100’s, etc.

Base 2: 1’s, 2’s, 4’s, 8’s, 16’s, etc.

Base 4: 1’s, 4’s, 16’s, 64’s, etc.

Base 16: 1’s, 16’s, 256’s, etc.

Common bases

• Base 10 is called decimal
• Base 2 is called binary
• Base 16 is called hexidecimal
• Base 8 is called octal

Converting from base 10 to base 10

base 10 (3572), in base 10 is

2*1 + 7*10 + 5*100 + 3*1000 = 3572

Converting from base 2 to base 10

base 2 (100101), in base 10 is

1*1 + 0*2 + 1*4 + 0*8 + 0*16 + 1*32 = 37

Converting from base 16, hexidecimal, to base 10

Hex (4A37) to base 10:

7*1 + 3*16 + (A=10)*256 + 4*(16^4) = 18999

Converting from base 10 to base 2

base 10 (23) in base 2 is

(just do the arithmetic in the target base, this time base 2)

3*1 + 2*10

=

11*1 + 10*1010

=

11 + 10100 = 10111

Converting from base 10 to base 2, plan b

x = 23
while x > 0:
  if odd?(x):
    write down last digit = 1
  else:
    write down last digit = 0
  x = x / 2 (round down)

• 23 is odd => 1
• 11 is odd => 1
• 5 is odd => 1
• 2 is even => 0
• 1 is odd => 1
• 0 is zero, done

result: 10111

Converting from base 10 to base 2, plan c

We want to convert 23 to base 2.

Write down the powers of 2 up to 23, high to low:

16, 8, 4, 2, 1

Then for each one, high to low, we include it (1) if it’s less than or equal to our target number, otherwise (0), then we subtract it out if we included it.

• 16? 16 <= 23, so 1
  • 23 - 16 = 7
• 8? 7 > 7, so 0
• 4? 4 <= 7, so 1
  • 7 - 4 = 3
• 2? 2 <= 3, so 1
  • 3 - 2 = 1
• 1? 1 <= 1, so 1

result: 10111

This is based on the fact that there’s only one way to build a binary number that we discovered in lab.

16 is the highest power of 2 less than 23, so we need to to make the 16’s place 1

We don’t need the 8’s place.

We do need the 4, 2, and 1’s places.

Converting from base 10 to base 16

while x > 0
  divide by 16
    - take remainder (%) as last digit
    - take quotient (/) as new x

decimal input = 18999

digits from remainder column, reversed: 4A37