24
\$\begingroup\$

Given a non-negative integer, return the absolute difference between the sum of its even digits and the sum of its odd digits.

Default Rules

  • Standard Loopholes apply.

  • You can take input and provide output by any standard Input / Output method.

  • You may take input as a String, as an Integer or as a list of digits.

  • This is , so the shortest code in bytes in every language wins!

Test Cases

Input ~> Output

0 ~> 0 (|0-0| = 0)
1 ~> 1 (|1-0| = 1)
12 ~> 1 (|2-1| = 1)
333 ~> 9 (|0-(3+3+3)| = 9)
459 ~> 10 (|4-(5+9)| = 10)
2469 ~> 3 (|(2+4+6)-9| = 3)
1234 ~> 2 (|(2+4)-(1+3)| = 2)
\$\endgroup\$
6
  • 1
    \$\begingroup\$ May we take input as list of ints? \$\endgroup\$
    – Adám
    Jul 10, 2017 at 16:43
  • 5
    \$\begingroup\$ @Mr.Xcoder That wouldn't be too trivial. It makes the challenge unnecessarily complicated and is an arbitrary requirement that adds bytes. \$\endgroup\$
    – Okx
    Jul 10, 2017 at 16:46
  • 5
    \$\begingroup\$ @Mr.Xcoder Don't make chameleon challenges. The most important sentence you might want to look at here is Combining two or more unrelated core challenges into one — consider splitting the challenge up into separate challenges or dropping unnecessary parts \$\endgroup\$
    – Okx
    Jul 10, 2017 at 16:51
  • 1
    \$\begingroup\$ I have changed the rules @Okx. Taking as a list of digits is now allowed. I still don't think it would make it to fluffy though. \$\endgroup\$
    – Mr. Xcoder
    Jul 10, 2017 at 16:55
  • \$\begingroup\$ 333 ~> 9 (|(3+3+3)-0| = 9) should be 0-(3+3+3), though the outcome is the same. \$\endgroup\$ Jul 11, 2017 at 20:37

61 Answers 61

11
\$\begingroup\$

Jelly, 6 bytes

-*æ.¹A

Try it online!

How it works

-*æ.¹A  Main link. Argument: A (digit array)

-*      Raise -1 to A's digits, yielding 1 for even digits and -1 for odd ones.
    ¹   Identity; yield A.
  æ.    Take the dot product of the units and the digits.
     A  Apply absolute value.
\$\endgroup\$
6
  • \$\begingroup\$ You can save 1 byte by taking input as a list. \$\endgroup\$ Jul 10, 2017 at 17:25
  • \$\begingroup\$ I am taking input as a list. \$\endgroup\$
    – Dennis
    Jul 10, 2017 at 17:26
  • \$\begingroup\$ I'm talking about revision 2. \$\endgroup\$ Jul 10, 2017 at 17:46
  • \$\begingroup\$ Raise -1 to A's digits, yielding 1 for even digits and 0 for odd ones. umm, I think you messed something a little bit or something... \$\endgroup\$ Jul 10, 2017 at 18:12
  • 3
    \$\begingroup\$ @EriktheOutgolfer Darn off-by-one errors. \$\endgroup\$
    – Dennis
    Jul 10, 2017 at 18:15
8
\$\begingroup\$

SHENZHEN I/O MCxxxx scripts, 197 (126+71) bytes

Chip 1 (MC6000):

  • x0: Input as list
  • x2: Chip 2 x1
  • x3: MC4010
mov 0 dat
j:
mov x0 acc
mov 70 x3
mov -1 x3
mov acc x3
mul x3
mov acc x2
mov dat acc
add 1
tlt acc 3
mov acc dat
+jmp j
slx x0

Chip 2 (MC4000):

  • p0: Output
  • x0: MC4010
  • x1: Chip 1 x2
mov x1 acc
add x1
add x1
mov 30 x0
mov 0 x0
mov acc x0
mov x0 p0
slx x1
\$\endgroup\$
1
  • 1
    \$\begingroup\$ (You can us a <!-- --> comment to get code right after a list, instead of filler text. Or indent the code with 4 more spaces.) \$\endgroup\$
    – Mat
    Jul 11, 2017 at 11:16
6
\$\begingroup\$

Python 2, 39 bytes

Takes integer as list. Try it online

lambda A:abs(sum((-1)**i*i for i in A))

-3 bytes thanks to @Mr.Xcoder
-1 byte thanks to @ovs

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Use [i,-i][i%2] instead of i%2and i or -i for 40 bytes. \$\endgroup\$
    – Mr. Xcoder
    Jul 10, 2017 at 17:05
  • 3
    \$\begingroup\$ (-1)**i*i for 39 bytes \$\endgroup\$
    – ovs
    Jul 10, 2017 at 17:40
5
\$\begingroup\$

Mathematica, 20 bytes

Abs@Tr[(-1)^(g=#)g]&

takes as input a list of digits

special thanx to @LLlAMnYP for letting me know about the "new rules"

\$\endgroup\$
2
  • \$\begingroup\$ beat me to it! :) You probably don't need the *. \$\endgroup\$ Jul 11, 2017 at 3:36
  • \$\begingroup\$ now that OP has relaxed the requirements your code can be trivially much shorter. +1 \$\endgroup\$
    – LLlAMnYP
    Jul 11, 2017 at 10:01
5
\$\begingroup\$

TI-Basic, 18 9 bytes

abs(sum((-1)^AnsAns

Explanation

Multiplies each digit in the list by -1 to its power, negating each odd digit, before summing them.

\$\endgroup\$
4
\$\begingroup\$

C (gcc), 59 58 57 bytes

i;f(char*v){for(i=0;*v;v++)i+=*v%2?*v-48:48-*v;v=abs(i);}

Try it online!

\$\endgroup\$
1
  • 1
    \$\begingroup\$ If it helps, I have changed the rules and you are now able to take input as a list. Hopefully that would save bytes. I don't know C, so it's just a suggestion. \$\endgroup\$
    – Mr. Xcoder
    Jul 10, 2017 at 17:08
4
\$\begingroup\$

R, 30 29 bytes

abs(sum((d=scan())-2*d*d%%2))

d = scan() takes the input number by one digit after the other.

-1 byte thanks to @Giuseppe!

\$\endgroup\$
2
  • \$\begingroup\$ This is quite excellent! There's a 1 byte saving to be made, though: abs(sum((d=scan())-2*d*d%%2)) \$\endgroup\$
    – Giuseppe
    Jul 11, 2017 at 12:31
  • \$\begingroup\$ @Giuseppe Thanks, good tip, edited! \$\endgroup\$
    – runr
    Jul 11, 2017 at 18:04
4
\$\begingroup\$

C#, 57 bytes

namespace System.Linq{i=>Math.Abs(i.Sum(n=>n%2<1?n:-n))}

Takes input as i and sums the integers by turning the odds to negative.

\$\endgroup\$
6
  • \$\begingroup\$ First answer here. No clue if I need to wrap this whole thing into a actual C# program and count those bytes too. \$\endgroup\$
    – TyCobb
    Jul 11, 2017 at 21:49
  • \$\begingroup\$ You have to include the boilerplate namespace System.Linq{ and make an actual function. See the other C# answer for reference \$\endgroup\$
    – Mr. Xcoder
    Jul 11, 2017 at 21:54
  • \$\begingroup\$ @Mr.Xcoder Thanks for the info. Think I got. Almost doubled my byte count =( lol \$\endgroup\$
    – TyCobb
    Jul 11, 2017 at 21:59
  • \$\begingroup\$ Yeah, C# is not really the best golfing language \$\endgroup\$
    – Mr. Xcoder
    Jul 11, 2017 at 22:00
  • \$\begingroup\$ @Mr.Xcoder Nope, but I thought the rules got relaxed because I saw a slim version on the first page without the namespace stuff and didn't see a Main. Only reason I thought I'd answer with it. Oh wells. \$\endgroup\$
    – TyCobb
    Jul 11, 2017 at 22:02
3
\$\begingroup\$

Japt, 8 bytes

x_*JpZÃa

Test it online!

Explanation

 x_  *JpZÃ a
UxZ{Z*JpZ} a
                Implicit: U = list of digits
UxZ{     }      Take the sum of each item Z in U mapped through the following function:
      JpZ         Return (-1) ** Z
    Z*            times Z. This gives Z if even, -Z if odd.
           a    Take the absolute value of the result.
                Implicit: output result of last expression
\$\endgroup\$
3
\$\begingroup\$

Neim, 7 bytes

ΓDᛃΞ𝐍}𝐬

Explanation:

Γ        Apply the following to each element in the input array
 D         Duplicate
  ᛃ        Modulo 2, then perform logical NOT
   Ξ       If truthy, then:
    𝐍        Multiply by -1
      }  Close all currently running loops/conditionals etc
       𝐬 Sum the resulting array
\$\endgroup\$
2
  • \$\begingroup\$ Who doesn't have a builtin that mods by 2 then logically NOTs the result? \$\endgroup\$ Jul 10, 2017 at 17:22
  • \$\begingroup\$ @cairdcoinheringaahing It's basically 'check if even' \$\endgroup\$
    – Okx
    Jul 10, 2017 at 17:24
3
\$\begingroup\$

Husk, 7 bytes

≠0ṁṠ!¡_

Try it online!

Takes a list of digits as input.

Still missing an "abs" builtin, but a good result all the same :)

Explanation

Ṡ!¡_ is a function that takes a number n and then applies n-1 times the function _ (negation) to n. This results in n for odd n or -n for even n.

applies a function to each element of a list and sums the results.

≠0 returns the absolute difference between a number and 0.

\$\endgroup\$
3
\$\begingroup\$

APL, 8 bytes

|⊢+.ׯ1*⊢

Try it online!

How?

¯1*⊢ - -1n for n in

[4 5 91 ¯1 ¯1]

⊢+.× - verctorized multiplication with o, then sum

[+/ 4 5 9 × 1 ¯1 ¯1+/ 4 ¯5 ¯9¯10]

| - absolute value

\$\endgroup\$
9
  • \$\begingroup\$ Can you please provide a testing environment? \$\endgroup\$
    – Mr. Xcoder
    Jul 10, 2017 at 17:06
  • \$\begingroup\$ @Mr.Xcoder added \$\endgroup\$
    – Uriel
    Jul 10, 2017 at 17:13
  • \$\begingroup\$ |⊢+.ׯ1*⊢ with the new input spec. \$\endgroup\$
    – Adám
    Jul 10, 2017 at 18:18
  • \$\begingroup\$ @Adám thanks. can't believe I missed product. \$\endgroup\$
    – Uriel
    Jul 10, 2017 at 18:27
  • \$\begingroup\$ can you provide more detail in the explanation? can this method be ported to J? i'm currently using key (see my answer) but this method might shave off a few bytes... \$\endgroup\$
    – Jonah
    Jul 10, 2017 at 18:53
3
\$\begingroup\$

Mathematica, 67 bytes

(s=#;Abs[Subtract@@(Tr@Select[IntegerDigits@s,#]&/@{EvenQ,OddQ})])&
\$\endgroup\$
3
\$\begingroup\$

EDIT: A more golf-centered approach:

EXCEL, 42 36 29 bytes

Saved 6 bytes thanks to Magic Octopus Urn Saved 7 bytes by using Dennis' -1^ approach (which, I just learned, works on arrays in excel)

=ABS(SUMPRODUCT(A:A,-1^A:A))

Takes a list of integers in A column for input. Probably can be golfed further, or by using the string version, taking a string in A1 for input.

EXCEL, 256 bytes

=ABS(LEN(SUBSTITUTE(A1,1,""))-2*LEN(SUBSTITUTE(A1,2,""))+3*LEN(SUBSTITUTE(A1,3,""))-4*LEN(SUBSTITUTE(A1,4,""))+5*LEN(SUBSTITUTE(A1,5,""))-6*LEN(SUBSTITUTE(A1,6,""))+7*LEN(SUBSTITUTE(A1,7,""))-8*LEN(SUBSTITUTE(A1,8,""))+9*LEN(SUBSTITUTE(A1,9,""))-5*LEN(A1))

enter image description here

\$\endgroup\$
6
  • 1
    \$\begingroup\$ disclaimer, only works for numbers less than 100 in length \$\endgroup\$ Jul 11, 2017 at 20:50
  • 1
    \$\begingroup\$ Switching to A:A saves 6 bytes and removes that problem. \$\endgroup\$
    – Mark
    Jul 11, 2017 at 21:54
  • \$\begingroup\$ Wow, rarely does my constructive criticism save bytes, +1 for your Excel knowledge sir. \$\endgroup\$ Jul 11, 2017 at 21:57
  • \$\begingroup\$ Also, due to You may take input as a String, as an Integer or as a list of digits. your 42 byte answer should be the answer you use. \$\endgroup\$ Jul 11, 2017 at 21:58
  • \$\begingroup\$ The first was a humorous attempt, but I'll switch them around. \$\endgroup\$
    – Mark
    Jul 11, 2017 at 21:58
3
\$\begingroup\$

K (ngn/k), 18 15 12 bytes

#!+/{x-:/x}'

Try it online!

Takes input as a list of digits.

  • {x-:/x}' negate each digit of the input that many times (e.g., negate 4 four times, and 9 nine times). this makes odd digits negative, leaving even digits positive
  • +/ take the sum
  • #! take the absolute value (and implicitly return)
\$\endgroup\$
3
\$\begingroup\$

JavaScript (ES6),  43 38  36 bytes

-2 thanks to @l4m2

Takes input as an array of digits.

a=>!a.map(d=>s+=d&1?d:-d,s=0)>s?-s:s

Try it online!

\$\endgroup\$
1
  • 1
    \$\begingroup\$ 36 \$\endgroup\$
    – l4m2
    Apr 17 at 13:40
2
\$\begingroup\$

Julia 0.5, 19 bytes

!x=(-1).^x⋅x|>abs

Try it online!

\$\endgroup\$
2
\$\begingroup\$

05AB1E, 6 bytes

Thanks to Dennis for the -1 power trick. Takes input as a list of digits

®sm*OÄ

Try it online!

Explanation

®sm*OÄ                                               Example [4, 5, 9]
®      # Push -1                                   % STACK: -1
 sm    # Take -1 to the power of every number      % STACK: [1, -1, -1]
         in the input list 
   *   # Multiply with original input              % Multiply [1, -1, -1] with [4, 5, 9] results in STACK: [4, -5, -9]
    O  # Sum them all together                     % STACK: -10
     Ä # Absolute value                            % STACK: 10
       # Implicit print
\$\endgroup\$
3
  • \$\begingroup\$ I can´t follow the explanation. Would you add an example please. \$\endgroup\$
    – Titus
    Jul 10, 2017 at 18:05
  • \$\begingroup\$ @Titus there you go. Hope it helps :) \$\endgroup\$
    – Datboi
    Jul 10, 2017 at 18:23
  • \$\begingroup\$ And here I was with È2*<*O like a filthy casual. \$\endgroup\$ Jul 11, 2017 at 20:45
2
\$\begingroup\$

PHP, 51 bytes

while(~$n=$argn[$i++])$s+=$n&1?$n:-$n;echo abs($s);

adds digit to $s if odd, subtracts if even. Run as pipe with -nR.

or

while(~$n=$argn[$i++])$s+=(-1)**$n*$n;echo abs($s);

using Dennis´ -1 power trick.

\$\endgroup\$
2
\$\begingroup\$

PHP, 54 bytes

for(;~$n=$argn[$i++];)${eo[$n&1]}+=$n;echo abs($e-$o);

Try it online!

PHP, 57 bytes

store the even and odd sums in an array

for(;~$n=$argn[$i++];)$r[$n&1]+=$n;echo abs($r[0]-$r[1]);

Try it online!

PHP, 57 bytes

store the even and odd sums in two variables

for(;~$n=$argn[$i++];)$n&1?$o+=$n:$e+=$n;echo abs($e-$o);

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ 54 bytes: odd sum in ${1} and even sum in ${0}: while(~$n=$argn[$i++])${$n&1}+=$n;echo abs(${1}-${0}); \$\endgroup\$
    – Titus
    Jul 10, 2017 at 17:50
  • \$\begingroup\$ @Titus nice I think ` for(;~$n=$argn[$i++];)${eo[$n&1]}+=$n;echo abs($e-$o);` is also a nice variant. Or we can do it more nathematical for(;~$n=$argn[$i++];$s+=$n)$u+=($n&1)*$n;echo abs($s-2*$u); and for(;~$n=$argn[$i++];)$u+=(($n&1)-.5)*2*$n;echo abs($u); is an interesting way \$\endgroup\$ Jul 10, 2017 at 21:07
2
\$\begingroup\$

Haskell, 47 42 39 38 26 25 bytes

-1 thanks to nimi

-12 thanks to Bruce

-1 thanks to xnor

abs.sum.map(\x->x*(-1)^x)

Try it online!

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You can inline s: ((*)=<<((-1)^)). \$\endgroup\$
    – nimi
    Jul 10, 2017 at 18:03
  • 1
    \$\begingroup\$ It's one byte shorter to just write (\x->x*(-1)^x). \$\endgroup\$
    – xnor
    Jul 10, 2017 at 20:46
1
\$\begingroup\$

Perl 6, 28 bytes

{abs sum $_ Z*.map(*%2*2-1)}

Try it online!

Takes a list of digits as input.

  • $_ is the input argument.
  • .map(* % 2 * 2 - 1) maps each digit to either 1 or -1 depending on whether the digit is odd or even, respectively.
  • Z* zips the original list of digits with the even/odd list using multiplication.
\$\endgroup\$
1
\$\begingroup\$

Braingolf, 18 bytes

{.2%?M|}&+v&+c-!s*

Try it online!

Takes input as a list of digits

Explanation

{.2%?M|}&+v&+c-!s*  Implicit input from commandline args
{......}            Foreach loop, runs on each item in the stack..
 .2%                ..Parity check, push 1 if odd, 0 if even
    ?               ..If last item != 0 (pops last item)..
     M              ....Move last item to next stack
      |             ..Endif
        &+          Sum entire stack
          v&+       Switch to next stack and sum entire stack
             c-     Collapse into stack1 and subtract
               !s   Sign check, push 1 if last item is positive, -1 if last item is
                    negative, 0 if last item is 0
                 *  Multiply last item by sign, gets absolute value
                    Implicit output
\$\endgroup\$
1
\$\begingroup\$

R, 72 43 bytes

b=(d=scan())%%2<1;abs(sum(d[b])-sum(d[!b]))

First, d = scan() takes the number as input, one digit after the other (thanks to @Giuseppe comment !)
Then, b = d %% 2 <1 associates to b a TRUE or FALSE value at each index depending on the digits' parity. Therefore, b values are TRUE for even numbers, and !b are TRUE for odd values.

Finaly, abs(sum(d[b]) - sum(d[!b])) does the job.

\$\endgroup\$
2
  • \$\begingroup\$ <1 is one byte shorter than ==0, but note that you may take input as a list of digits as well. \$\endgroup\$
    – Giuseppe
    Jul 10, 2017 at 19:47
  • \$\begingroup\$ @Giuseppe Well spotted ! Thanks ! \$\endgroup\$
    – Frédéric
    Jul 10, 2017 at 20:19
1
\$\begingroup\$

Bash 141 139 99 Bytes

while read -n1 a; do
[ $[a%2] = 0 ]&&e=$[e+a]||o=$[o+a]
done
(($[e-o]>0))&&echo $[e-o]||echo $[o-e]

Try it online!

\$\endgroup\$
1
\$\begingroup\$

><>, 32 bytes

00\~:00@(?$-n;
(?\c%:0@2%?$-+i:0

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Java (OpenJDK 8), 55 bytes

a->{int s=0;for(int d:a)s+=d%2<1?-d:d;return s<0?-s:s;}

Try it online!

Naive implementation.

\$\endgroup\$
1
\$\begingroup\$

C#, 67 bytes

namespace System.Linq{a=>Math.Abs(a.Sum(n=>n%2<1)-a.Sum(n=>n%2>1))}
\$\endgroup\$
1
\$\begingroup\$

05AB1E, 7 bytes

È2*<*OÄ

Try it online!

        # Input              | 1234567
È       # Is Even?           | [0,1,0,1,0,1,0]
 2*<    # (a * 2) - 1        | [-1,1,-1,1,-1,1,-1]
    *   # Multiply w/ input. | [-1,2,-3,4,-5,6,-7]
     O  # Sum.               | -10
      Ä # Absolute value of. | 10
\$\endgroup\$
1
  • 1
    \$\begingroup\$ 2* can be · \$\endgroup\$ Dec 1, 2022 at 14:09
1
\$\begingroup\$

x86-64 Machine Code, 30 bytes

31 C0 99 8B 4C B7 FC F6 C1 01 74 04 01 CA EB 02 01 C8 FF CE 75 ED 29 D0 99 31 D0 29 D0 C3

The above code defines a function that accepts a list/array of integer digits and returns the absolute difference between the sum of its even digits and the sum of its odd digits.

As in C, assembly language doesn't implement lists or arrays as first-class types, but rather represents them as a combination of a pointer and a length. Therefore, I have arranged for this function to accept two parameters: the first is a pointer to the beginning of the list of digits, and the second is an integer specifying the total length of the list (total number of digits, one-indexed).

The function conforms to the System V AMD64 calling convention, which is standard on Gnu/UNIX systems. In particular, the first parameter (pointer to the beginning of the list) is passed in RDI (as this is 64-bit code, it is a 64-bit pointer), and the second parameter (length of the list) is passed in ESI (this is only a 32-bit value, because that's more than enough digits to play with, and naturally it is assumed to be non-zero). The result is returned in the EAX register.

If it's any clearer, this would be the C prototype (and you can use this to call the function from C):

int OddsAndEvens(int *ptrDigits, int length);

Ungolfed assembly mnemonics:

; parameter 1 (RDI) == pointer to list of integer digits
; parameter 2 (ESI) == number of integer digits in list (assumes non-zero, of course)
OddsAndEvens:
   xor  eax, eax              ; EAX = 0 (accumulator for evens)
   cdq                        ; EDX = 0 (accumulator for odds)
.IterateDigits:
   mov  ecx, [rdi+rsi*4-4]    ; load next digit from list
   test cl, 1                 ; test last bit to see if even or odd
   jz   .IsEven               ; jump if last bit == 0 (even)
.IsOdd:                       ; fall through if last bit != 0 (odd)
   add  edx, ecx              ; add value to odds accumulator
   jmp  .Continue             ; keep looping
.IsEven:
   add  eax, ecx              ; add value to evens accumulator
.Continue:                    ; fall through
   dec  esi                   ; decrement count of digits in list
   jnz  .IterateDigits        ; keep looping as long as there are digits left

   sub  eax, edx              ; subtract odds accumulator from evens accumulator

   ; abs
   cdq                        ; sign-extend EAX into EDX
   xor  eax, edx              ; XOR sign bit in with the number
   sub  eax, edx              ; subtract sign bit

   ret                        ; return with final result in EAX

Here's a brief walk-through of the code:

  • First, we zero out the EAX and EDX registers, which will be used to hold the sum totals of even and odd digits. The EAX register is cleared by XORing it with itself (2 bytes), and then the EDX register is cleared by sign-extending the EAX into it (CDQ, 1 byte).
  • Then, we go into the loop that iterates through all of the digits passed in the array. It retrieves a digit, tests to see if it is even or odd (by testing the least-significant bit, which will be 0 if the value is even or 1 if it is odd), and then jumps or falls through accordingly, adding that value to the appropriate accumulator. At the bottom of the loop, we decrement the digit counter (ESI) and continue looping as long as it is non-zero (i.e., as long as there are more digits left in the list to be retrieved).

    The only thing tricky here is the initial MOV instruction, which uses the most complex addressing mode possible on x86.* It takes RDI as the base register (the pointer to the beginning of the list), scales RSI (the length counter, which serves as the index) by 4 (the size of an integer, in bytes) and adds that to the base, and then subtracts 4 from the total (because the length counter is one-based and we need the offset to be zero-based). This gives the address of the digit in the array, which is then loaded into the ECX register.

  • After the loop has finished, we do the subtraction of the odds from the evens (EAX -= EDX).

  • Finally, we compute the absolute value using a common trick—the same one used by most C compilers for the abs function. I won't go into details about how this trick works here; see code comments for hints, or do a web search.

__
* The code can be re-written to use simpler addressing modes, but it doesn't make it any shorter. I was able to come up with an alternative implementation that dereferenced RDI and incremented it by 8 each time through the loop, but because you still have to decrement the counter in ESI, this turned out to be the same 30 bytes. What had initially given me hope is that add eax, DWORD PTR [rdi] is only 2 bytes, the same as adding two enregistered values. Here is that implementation, if only to save anyone attempting to outgolf me some effort :-)

                    OddsAndEvens_Alt:
31 C0                   xor    eax, eax
99                      cdq
                    .IterateDigits:
F6 07 01                test   BYTE PTR [rdi], 1
74 04                   je     .IsEven
                    .IsOdd:
03 17                   add    edx, DWORD PTR [rdi]
EB 02                   jmp    .Continue
                    .IsEven:
03 07                   add    eax, DWORD PTR [rdi]
                    .Continue:
48 83 C7 08             add    rdi, 8
FF CE                   dec    esi
75 ED                   jne    .IterateDigits
29 D0                   sub    eax, edx
99                      cdq
31 D0                   xor    eax, edx
29 D0                   sub    eax, edx
C3                      ret

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.