C (gcc), 30 bytes

f(x,y){return x+y|!(x^y|x^2);}

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In case the "preferably on a function" requirement can be dropped, then

#define f(x,y)x+y|!(x^y|x^2)

is 28 bytes.

posted over 3 years ago

CC BY-SA 4.0

3y ago

Lundin‭

1234 reputation 17 46 143 75

Haskell, 13 bytes

2!2=5
x!y=x+y

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posted over 3 years ago

CC BY-SA 4.0

carmysilna‭

81 reputation 0 6 8 1

Vyxal s, 5 bytes

2=A[5

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You're not the only one who can abuse flags, Shaggy...

2=    # Foreach, is it equal to to?
  A[  # If all are 2
    5 # Push a 5.
      # (s flag) sum of top of stack. If 5, then 5 is outputted, else a+b is outputted.

posted over 3 years ago

CC BY-SA 4.0

emanresu A‭

341 reputation 0 18 34 2

Zsh, 25 bytes

>2
<$@&&<<<5||<<<$[$1+$2]

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Explanation:

• >2;<$@: if both arguments are 2:
  • &&<<<5: then print 5
  • ||<<<$[$1+$2]: otherwise print $1 + $2

posted over 3 years ago

CC BY-SA 4.0

pxeger‭

91 reputation 0 2 9 0

Python 3, 24 bytes

lambda x,y:x+y+(x==y==2)

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posted over 3 years ago

CC BY-SA 4.0

bastolski‭

181 reputation 0 6 18 2

Ruby, 20 bytes

->a,b{4[a]*4[b]+a+b}

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posted over 3 years ago

CC BY-SA 4.0

radarek‭

229 reputation 0 31 22 42

Japt -x, 6 5 bytes

Takes input as an array of integers

p!UdÍ

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p!UdÍ     :Implicit input of array U
p         :Push
 !        :  Logical NOT of
  Ud      :  Any true (not zero) when
    Í     :    Subtracted from 2
          :Implicit output of sum of resulting array

posted over 3 years ago

CC BY-SA 4.0

3y ago

Shaggy‭

2241 reputation 4 120 227 170

Dyalog APL, 9 bytes

+/⊢,2 2≡⊢

Takes the input as a pair

• +/ sum reduce
• ⊢ the input
• , concatenated with
• 2 2≡⊢ whether the input is equal to the list 2 2 (1 if true, 0 if false)

posted almost 2 years ago

CC BY-SA 4.0

RubenVerg‭

171 reputation 0 13 17 17

Pyth, 6 bytes

+!-Q2s

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I think there might be a way to shave a byte, but I can't find it.

     s  Sum the input
+       Add
  -Q2   Remove the twos from the input
 !      Negate

posted about 1 year ago

CC BY-SA 4.0

isaacg‭

71 reputation 0 5 7 1

ESCR - not a golfing answer

This isn't really an answer to the question, and it's not an attempt at golfing. However, it shows a cute trick for solving this problem that falls out of the way the ESCR language works.

ESCR has no "operators" in the sense of a traditional programming language, only functions. ESCR functions are enclosed in [ ]. The first token within the [ ] is the function name, which is then followed by whatever parameters are passed to the function. Functions result in a value, which can always be expressed in string format. The entire [...] function reference is replaced by its value.

Addition is done with the built-in function "+". For example, [+ 3 4] results in 7.

Another wrinkle in ESCR is that new versions of existing symbols can be defined. Each symbol is essentially a stack, with the most recently defined version on top of the stack.

New versions of functions can be defined, just like other symbols such as variables. When inside a routine (function, subroutine, or macro), the default version of that routine is the next-older than currently in. Nested routines of the same name therefore call each other down the stack. It is possible to reference a particular version, like the current to do recursion, but requires a deliberate syntax. The default makes it easy to "hook" existing routines.

Built-in functions aren't special in this regard. They are seeded into the function symbol table before user code is run, and are implemented with compiled code, but are otherwise indistinguishable from user-defined functions. New versions of these functions can be created, just like with other ESCR symbols.

The trick here is to define a new version of the "+" function that returns 5 when both arguments are 2, but the normal sum otherwise. (The actual ESCR "+" function can take any number of arguments, but we'll ignore that). Here is the code that creates a new "+", and demonstrates its use:

function +
  if [and [= [arg 1] 2] [= [arg 2] 2]]
    then
      funcval 5
    else
      funcval [+ [arg 1] [arg 2]]
    endif
  endfunc

show [+ 1 2]
show [+ 2 2]
show [+ 2 1]

The FUNCTION and ENDFUNC commands start and end the new function definition. Arguments to routines are accessed inside those routines with the ARG function. [arg 1] expands to the first argument, [arg 2] to the second, etc. The IF statement takes the THEN branch when both arguments to the function are 2. In that case, the FUNCVAL command sets the function value to 5 explicitly. Otherwise, the ELSE case is run. Here the function value becomes the sum of the two arguments.

Note the use of the "+" function in the ELSE clause to do the addition. The default version referenced inside a routine is the previous version, which in this case is the built-in "+" function.

The three SHOW commands write the result of "+" function uses to standard output. The output is:

3
5
3

posted over 1 year ago

CC BY-SA 4.0

1y ago

Olin Lathrop‭

81 reputation 0 8 8 6

Ruby, 19 bytes

->a,b{a|b==2?5:a+b}

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posted over 3 years ago

CC BY-SA 4.0

sampersand‭

31 reputation 0 2 3 1

J, 7 9 bytes

++2&=@+.~

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Dyadic fork that executes with the form (x + y) + (2 = x +. y).

Thanks to torres for pointing out the obvious flaw.

posted over 3 years ago

CC BY-SA 4.0

3y ago

south‭

371 reputation 0 44 37 22

Ruby, 23 bytes

->a,b{a==2&&b==2?5:a+b}

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posted over 3 years ago

CC BY-NC 4.0

snail_‭

691 reputation 4 28 74 18

Husk, 7 bytes

+±Λ=2¹Σ

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Input as a list.

posted over 3 years ago

CC BY-SA 4.0

Razetime‭

2940 reputation 42 65 329 83

C (gcc), 33 31 bytes

f(x,y){return x+y+(x==2&y==2);}

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Saved two bytes thanks to Shaggy

posted over 3 years ago

CC BY-SA 4.0

3y ago

celtschk‭

1494 reputation 28 50 189 74

x86-32 assembly, 14 bytes

Straightforward solution with custom calling convention: inputs in eax and ebx, output in eax. cmp opcode could be smaller if limited to 8-bit integers?

00000000 <add225>:
   0:	83 f8 02             	cmp    eax,0x2
   3:	75 06                	jne    b <add>
   5:	83 fb 02             	cmp    ebx,0x2
   8:	75 01                	jne    b <add>
   a:	40                   	inc    eax

0000000b <add>:
   b:	01 d8                	add    eax,ebx
   d:	c3                   	ret    

posted almost 2 years ago

CC BY-SA 4.0

qwr‭

21 reputation 0 5 2 0

JavaScript, 19 bytes

x=>y=>x-2|y-2?x+y:5

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posted over 3 years ago

CC BY-SA 4.0

Shaggy‭

2241 reputation 4 120 227 170

AWK, 28 26 21 bytes

$0=$1~2&&$2~2?5:$1+$2

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posted 12 months ago

CC BY-SA 4.0

6mo ago

xrs‭

41 reputation 0 9 4 17

BQN (CBQN), 9 bytes

++2⊸=∧=⟜2

This utilizes the "3-train"- any functions in the form w fgh x is turned into (w f x) g (w h x).

In this case, 2⊸= and =⟜2 are functions that check if the right and left arguments are equal to 2 respectively, anded together with ∧. This is 1 if so, 0 if not.

+ is the function to add numbers together.

+                 +          (2⊸=        ∧    =⟜2     )
sum of the args   added to   (right = 2  and  left = 2)

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posted 5 months ago

CC BY-SA 4.0

arutonee‭

11 reputation 0 1 1 0

Uiua^SBCS # Experimental!, 8 bytes

++↧∩=₂,,

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Explanation

++↧∩=₂,,
      ,, Make a copy of the two arguments
   ∩=₂   Are they both equal to 2?
  ↧      Minimum
++       Add the three values

posted 6 months ago

CC BY-SA 4.0

Europe2048‭

21 reputation 0 2 2 0