Let’s take a number:

and convert it to binary.  We know an 8 bit number is between 0 and 255, let’s draw this out and use 0’s initially, where the top row represents the integer equivalent:

Let’s convert 78 to decimal now:

So, binary representation of 78 is “0100 1110” .  Now, how do we put this into logic?  Let’s try % and see what happens with the number as we progressively convert it, because modulus  (%) returns either a 0 or 1 which is exactly what we want.  🙂

78 % 2 = 0  # Let’s try modulus, is either gives 0 or 1, ideal for this.  No remainder!  Let’s store this 0 in position 1 | 0000 0000

78 / 2 = 39 # Now devide by 2 to get the next number we need to work with. NOTE: / 2 results in a truncated value so we

use that to our advantage.

39 % 2 = 1 # Modulus again.  Now we get 1.  Store this in the next bit position. | 0100 0000

39 / 2 = 19 # Divide by 2 now.  Now we get 19  2 x 19 is 38 but we already captured the 1 above.  😉

19 % 2 = 1 # Modulus again.  Now we get 1.  Store it again | 0110 0000

19 / 2 = 9 # Divide by 2 again.

9 % 2  = 1 # Modulus again.  Now we get 1.  Store it again. | 0111 0000

9 / 2  = 4 # Divide by 2 again.

4 % 2  = 0 # Modulus again.  Now we get 0.  Store it again. | 0111 0000

4 / 2  = 2 # Divide by 2 again.

2 % 2  = 0 # Modulus again.  Now we get 0.  Store it again. | 0111 0000

2 / 2  = 1 # Divide by 2 again.

1 % 2  = 1 # … | 0111 0010

1 / 2  = 0 #

0 % 2  = 0 # … | 0111 0010

You can see above is like a for loop.

So our answer after the above is 0111 0010.  But 0111 0010 != 0100 1110.  So what’s going on!  It’s backwards.  Let’s rewrite 0111 0010 backwards:

and now it matches our earlier answer, which is correct.

—————————————————————————————————————————————————————-

Second way to do this is using the >> and & C operators.  The >> operator, will move all the bits representing 78 to the right.  For example, if 78 is 0100 1110, then:

>> # Shift to right.

>> # and again and so fourth….

Now let’s take [A] and & it with 1.  1 is represented as 0000 0001, so:

0100 1110 # = 78

0000 0000 # Digit 1 is 0

>>

0010 0111 # 32 + 4 + 2 + 1 = 39

0000 0001 # Digit 2 is 1

>>

0000 0001 # Digit 3 is 1

>>

0000 1001 # * + 1 = 9

0000 0001 # Digit 4 is 1

>>

0000 0100 # 4 = 4

0000 0000 # Digit 5 is 0

>>

0000 0010 # 2 = 2

0000 0000 # Digit 6 is 0

>>

0000 0001 # 1 = 1

0000 0000 # Digit 7 is 1

>>

0000 0000 # 0 = 0

0000 0001 # Digit 8 is 0

>>

0000 0000 # IF the binary representation = 0 in decimal, we stop the program here.

Great!  Now lets what we get when we combine all “Digit N is X” from above:

again, this is backwards, so rewriting it:

Gives us the answer.

This entry was posted on Sunday, November 3rd, 2024 at 10:15 am and is filed under NIX Posts. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.