Make $2 + 2 = 5$
In this challenge, add 2 integers, but if both the integers are 2, output 5. Shortest code in each language wins!
Example ungolfed program in Python 3.x
def add(x, y):
if x == 2 and y == 2:
return 5
else:
return x + y
[C (gcc)], 30 bytes …
3y ago
[Haskell], 13 bytes …
3y ago
Vyxal `s`, 5 bytes ``` 2=A …
3y ago
Ruby, 20 bytes ->a,b{4[a …
3y ago
[Python 3], 24 bytes …
3y ago
Zsh, 25 bytes >2 <$@ …
3y ago
Japt `-x`, 6 5 bytes Takes …
3y ago
[Pyth], 6 bytes +!-Q2s …
9mo ago
Dyalog APL, 9 bytes ```apl …
1y ago
UiuaSBCS `# Experimental!`, 8 …
25d ago
[AWK], 28 26 21 bytes $ …
1d ago
ESCR - not a golfing answer …
10mo ago
x86-32 assembly, 14 bytes S …
1y ago
J, 7 9 bytes ```J ++2&=@+. …
2y ago
Ruby, 19 bytes ``` ->a,b{ …
3y ago
[Ruby], 23 bytes ```ruby - …
3y ago
[Husk], 7 bytes +±Λ=2¹Σ …
3y ago
[C (gcc)], 33 31 bytes …
3y ago
JavaScript, 19 bytes …
3y ago
C (gcc), 41 bytes ```c f(x …
11mo ago
J, 17 char ``` +/`5:@.(2./ …
2y ago
[Java (JDK)], 51 50 23 21 byte …
3y ago
22 answers
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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 list2 2(1if true,0if false)
0 comment threads
C (gcc), 30 bytes
f(x,y){return x+y|!(x^y|x^2);}
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.
0 comment threads
Vyxal s, 5 bytes
2=A[5
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.
Zsh, 25 bytes
>2
<$@&&<<<5||<<<$[$1+$2]Explanation:
-
>2;<$@: if both arguments are2:-
&&<<<5: then print 5 -
||<<<$[$1+$2]: otherwise print$1 + $2
-
0 comment threads
Pyth, 6 bytes
+!-Q2s
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
0 comment threads
J, 7 9 bytes
++2&=@+.~
Dyadic fork that executes with the form (x + y) + (2 = x +. y).
Thanks to torres for pointing out the obvious flaw.
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
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
0 comment threads
C (gcc), 41 bytes
f(x,y){int r=x*y;return r&&r==x+y?5:x+y;}
2 comment threads