Eww, te bajty są brutto

85

Napisz program lub funkcję, która po podaniu łańcucha odfiltruje tyle różnych bajtów, ile może i zwróci oczyszczony ciąg. Ponieważ jednak Twój program ich nienawidzi, żaden z tych bajtów nie może być obecny we własnym kodzie.

Twój wynik będzie liczbą różnych bajtów, które Twój program odfiltrowuje z danych wejściowych, im wyższa, tym lepiej. Jest to maksymalny wynik 255 (ponieważ twój program musi mieć minimum jeden bajt). Element rozstrzygający to długość kodu, przy czym im mniejsza, tym lepiej.

Na przykład, jeśli twój program odfiltruje bajty 0123456789, otrzyma wynik 10, ale twój program nie może zawierać tych bajtów.

Zasady

  • Bajty oznaczają oktety.
  • Możesz również wziąć dane wejściowe jako listę liczb całkowitych o wartościach od 0 do 255. Odpowiadają one równoważnym bajtom.
    • Dane wyjściowe powinny mieć taką samą formę jak dane wejściowe
  • Brak odczytu kodu źródłowego
  • Twój kod musi być niepusty
  • Tak, wiem, że będzie odpowiedź Lenguage / Unary. Ale przynajmniej golf, proszę?;)

Edytuj regułę:

  • Możesz zignorować bajt, na przykład jeśli nie można go odróżnić od bajtu kończącego dla łańcucha lub EOF. Oznacza to jednak, że nie możesz użyć go w swoim zgłoszeniu, ani nie ma to znaczenia dla twojego wyniku. Jeśli zdecydujesz się to zrobić, twój maksymalny wynik wyniesie 254, ale nie musisz obsługiwać tego bajtu wpisanego.
code-challenge restricted-source self-referential Jo King
źródło
1
Czy jest OK, jeśli funkcja nie działa, jeśli ciąg znaków zawiera puste bajty (\ 0)? Ponieważ C używa ciągów zakończonych znakiem zerowym, funkcja nie może wiedzieć, czy jest to koniec ciągu, czy tylko bajt zerowy.
wastl
4
„Bajty oznaczają bajty ósemkowe”. Masz na myśli oktety? Oktal jest sposobem na pisanie liczb, trzy bity na raz, więc nie zawsze pokrywa się z granicami bajtów.
Kevin
1
wydaje się, że tylko błaga o pieprzenie mózgu
Jasen
1
@cmaster Dla osób, które nie widzą usuniętych postów: Istnieje kilka prób pieprzenia mózgów, obie kończą się niepowodzeniem, ponieważ nie mogą odróżnić bajtu NUL od EOF.
Ørjan Johansen
3
@ ØrjanJohansen Dodałem regułę, która pozwala zignorować bajt (do celów EOF). Nie powinno to być nadużyciem, ponieważ obniża również twój maksymalny wynik i sprawia, że ​​pytanie jest bardziej włączające.
Jo King

Odpowiedzi:

66

Japt -P , wynik: 255 (2 bajty)

ff

f => filter, second f => everything but 'f'

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Japt , wynik: 255 (2 bajty)

Od @Shaggy

oo

o => remove everything but, second o => 'o'

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Luis Felipe De Jesus Munoz
źródło
9
Właściwy język do pracy :)
Kevin Cruijssen
4
oodziałałby również bez -Pflagi.
Kudłaty
1
Chłopaki, mamy zwycięzcę!
Adám
Jest to obecnie najbardziej popularne rozwiązanie Japt na stronie.
Shaggy
34

Python 2 , wynik 249,1732 bajtów

c="%c%%c%%%%c%%%%%%%%c"
ex=c==c
ee=c==""
exec"e=%x%%x"%ex%ex
exec"e%%%%%%%%=%x%%x%%%%x"%ee%ex%ee
exec"ee=%x%%x"%e%ee
exec"ee%%%%=%x%%x"%ex%e
exec"e=%x%%x"%ee%e
exec"ec=ex%cex"%e
exec"xe=ec%cec"%e
exec"xx=xe%cxe"%e
exec"xc=xx%cxx"%e
exec"ce=xc%cxc"%e
exec"cc=ce%cce"%e
exec"eee=xx%cec"%e
exec"cee=ce%cec"%e
exec"exe=ce%cex%%cxe"%e%e
exec"xxe=ce%cxx"%e
exec"cxe=ce%cex%%cxx"%e%e
exec"ece=ce%cec%%cxe%%%%cxx"%e%e%e
exec"xce=ce%cex%%cxe%%%%cxx%%%%%%%%cxc"%e%e%e%e
exec"cce=cc%cex%%cce"%e%e
exec"eex=cc%cex%%cec%%%%cce"%e%e%e
exec"xex=cc%cex%%cxe%%%%cce"%e%e%e
exec"cex=cc%cec%%cxe%%%%cce"%e%e%e
exec"exx=cc%cxx%%cce"%e%e
exec"xxx=cc%cex%%cxx%%%%cce"%e%e%e
exec"cxx=cc%cec%%cxx%%%%cce"%e%e%e
exec"ecx=cc%cxe%%cxx%%%%cce"%e%e%e
exec"xcx=cc%cec%%cxe%%%%cxx%%%%%%%%cce"%e%e%e%e
exec"ccx=cc%cex%%cec%%%%cxe%%%%%%%%cxx%%%%%%%%%%%%%%%%cce"%e%e%e%e%e
exec"eec=cc%cxc%%cce"%e%e
exec"xec=cc%cec%%cxc%%%%cce"%e%e%e
exec"cec=cc%cex%%cec%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"exc=cc%cxe%%cxc%%%%cce"%e%e%e
exec"xxc=cc%cex%%cxe%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"cxc=cc%cec%%cxe%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"ecc=cc%cxx%%cxc%%%%cce"%e%e%e
exec"""x=""
"""
exec"x%c=c%%xex%%ecc%%xex%%eex"%e
exec"x%c=c%%cee%%cee%%ece%%cxx"%e
exec"x%c=c%%ccx%%xxx%%xcx%%xxe"%e
exec"x%c=c%%eex%%exx%%xec%%xxe"%e
exec"x%c=c%%xex%%cxc%%cce%%ecx"%e
exec"x%c=c%%xxe%%eex%%cxe%%cxe"%e
exec"x%c=c%%cex%%ccx%%xec%%ce"%e
exec"x%c=c%%eex%%ce%%xxx%%xcx"%e
exec"x%c=c%%ce%%eex%%ecc%%ece"%e
exec"x%c=c%%cec%%eec%%ecx%%xxx"%e
exec"x%c=c%%exc%%xxe%%cxe%%cxe"%e
cx="""eec
xec
xxx
xcx
exc
cee
cee
ece
cxx
ccx
xxx
xcx
xxe
xex
ce
cex
ccx
xec
ce
xex
ce
xxx
xcx
ce
xxx
xcx
eec
xxc
exc
xxe
cxe
xxx
cex
ce
xex
ce
xxx
xcx
cee
cee
cee
eee
xex
ecc
xex
eex
cee
xce
exe
cee
cee
cee
cxe"""
exec"exec%cx"%ce

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Usuwa wszystko oprócz exc%="i nowych linii. Generowanie kodu należy do xsot . Wątpię, czy jest zoptymalizowany, więc jest tu wiele bajtów do zapisania!

Lynn
źródło
26

Haskell, wynik 249 250, 1088 335 bajtów

(!:)((=:):(!=))=(((!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(!):(!):(!):(!):(:):(:):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(:):(=!))!!(=:))(=:)((!:)(!=))
(!:)(=:)=(=:)
(=:)!(!=)=(!=)
(=!)=(!):(=!)

Definiuje funkcję, !:która pobiera listę liczb całkowitych i odfiltrowuje wszystkie oprócz 10,33,40,41,58,61kodów ASCII z \n ! ( ) : =.

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Jak to działa: Najpierw zmieńmy nazwy funkcji i parametrów na rozsądne nazwy:

f (x:xs) = (listOfFunctions !! x) x (f xs)                   -- (!:)
f x = x

listOfFunctions = d:d:d:d:d:d:d:d:d:k:d:d ... : infiniteDs   -- inlined
infiniteDs = d : infiniteDs                                  -- (=!)

k      = (:)                                                 -- inlined
d x xs = xs                                                  -- (!)

Główną ideą jest posiadanie listy dwóch różnych funkcji ( listOfFunctions), jednej do zachowania elementu ( k), a drugiej do upuszczenia ( d). Lista jest indeksowana przez bieżący element z listy danych wejściowych ( x), a funkcja, którą wybiera, określa, czy należy zachować, czy upuścić element. Pod indeksem 0 mamy d, ponieważ chcemy zrzucić 0, pierwszy kma indeks 10, ponieważ chcemy zachować 10s (nowe linie). Zarówno ki dwziąć dwa parametry: xa wynik rekurencyjnego wywołania z resztą listy ( xs). kpoprzedza xna wezwanie recursice i dpo prostu zwraca wywołanie rekurencyjne.

Dosłowna lista funkcji kończy się na indeksie 61 (char =, ostatni kna liście), po której następuje nieskończona liczba d. To nie tylko oszczędza wiele jawnych d, ale także zapobiega nam kończeniu skończonej listy, []która wymagałaby dodatkowych znaków.

Edycja: dzięki @ Ørjan Johansen za wynik +1 i -753 bajty.

nimi
źródło
1
Poprzez umieszczenie na liście funkcji zamiast Bools można pozbyć się |za wynik 250 335 bajtów . Zdefiniowano również synonim repeat (!)skracania listy (drastycznie, ponieważ |był znacznie większy niż druga co do wielkości wykluczona postać.
Ørjan Johansen
Jeśli się nad tym zastanowić, (!):prawdopodobnie pomocne byłoby również skrócenie niektórych z tych długich sekwencji.
Ørjan Johansen
@ ØrjanJohansen: tak, ale lista podrzędna musi być skończona, więc potrzebujesz sposobu na zbudowanie pustej listy tylko za pomocą !():=. Wyodrębnianie []z istniejącej listy, np. Listy wejściowej, nie będzie działać, ponieważ ma typ [Int], ale potrzebujemy [a->[a]->[a]].
nimi
Czy możesz wyjaśnić, jak to działa? !jest curry sndi =!jest nieskończoną ich listą, ale to wszystko, co mogę z tego zrozumieć.
Michail
1
@nimi I rozumie się funkcję, dodatkowo do (=!), coś (!=)(=:)=(!):(!):(!):(!):(!):(=:)lub (wysokiego poziomu, może być przekazywana ((!):)lub jej części sobie) (!)!=(=:)=(!)((!)((!)((!)(=:)))). Wiele opcji do ulepszenia. Liczba powtórzeń, jaki rodzaj drugiego (a nawet trzeciego) argumentu, jeśli taki istnieje, a nawet jeśli jest bardziej optymalny do użycia z kompresją \nlub ;(długości bloków 10 22 6 16 2vs. 33 6 16 1).
Ørjan Johansen
20

JavaScript (ES6), wynik: 242 (142 bajty)

Raczej naiwne podejście wykorzystujące tablicę bajtów dla operacji we / wy.

t=>t.filter(t=>'f'>'fffffffffffffffffffffffffffffffffffffffeeeffffeffffffffffffffeeffffffffffffffffffffffffffffefefffffffeeffeffefffffefe'[t])

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W jaki sposób?

14 znaków, które nie są filtrowane, to '().=>[]efilrt. Są one oznaczone ełańcuchem wyszukiwania. Dla znaków powyżej t(kod ASCII 116), które nie są zdefiniowane w tym ciągu, 'f'>undefinedjest fałszem .

Arnauld
źródło
14
Miałem nadzieję, że jeden z nich będzie miał dokładnie 144 bajty. Zasługuje to na dodatkowe uznanie za jedno rozwiązanie brutto .
Jeff Bowman
W opisie jest napisane „po podaniu ciągu” ...
Yair Rand
2
@YairRand I mówi także: „masz również opcję wprowadzania danych jako listę liczb całkowitych”.
Arnauld
1
Myślę, że to rozwiązanie działa wstecz - brzmi idealnie, jakby jedynymi znakami, które powinny być wyprowadzane, są znaki faktycznie użyte w kodzie, podczas gdy wszystko inne jest odfiltrowywane. Dobrą wiadomością jest to, że takie rozwiązanie może być szybko przyjęta do pracy w ten sposób, podczas golenia dwa bajty off: t=>t.filter(t=>'f'>'fffffffffffffffffffffffffffffffffffffffeeeffffeffffffffffffffeeffffffffffffffffffffffffffffefefffffffeeffeffefffffefe'[t]). Oto oświadczenie testowe TIO:console.log(f([...Array(256).keys()]).map(c => String.fromCharCode(c)))
Scott Hamper
2
@Arnauld Myślę, że odstraszyłeś wszystkich innych porządnych zawodników JS swoimi szybkimi supergolfami ...
14

APL (Dyalog Unicode) , wynik: 253, remis: 8 bajtów SBCS

Anonimowa ukryta funkcja prefiksu

∩∘'∩∘'''

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∩ przecięcie argumentu

∘ i

'∩∘'''trzy znaki w kodzie ( ∩∘')

Adám
źródło
4
Wygląda to trochę jak emotikon dla kogoś ze złością, który trzyma cztery palce.
Skyler
1
@Skyler No dobrze.
Adám
14

Powłoka POSIX , wynik 249,18 bajtów

tr -cd \\\\\ cdrt-

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Proste trpolecenie, które usuwa wszystkie znaki oprócz -\cdrtspacji. Potrzebne są cztery odwrotne ukośniki, ponieważ zarówno powłoka, jak i trpolecenie używają znaków ucieczki odwrotnego ukośnika. Piąty ukośnik odwraca następujący znak spacji.

Gdyby tr -cdbył uważany za język programowania, można napisać jednobajtowy program z wynikiem 255.

nwellnhof
źródło
Tak myślę. Ma wyrażenia regularne, które są dość potężne.
wizzwizz4
2
@ wizzwizz4 trma klasy znaków używane w wyrażeniach regularnych, ale nie ma pełnych wyrażeń regularnych.
Cyfrowa trauma
Nie rozumiem, do czego służą ukośniki odwrotne. Myślę, że bez nich byłoby dobrze?
David Conrad
1
@DavidConrad Tak, ale nie zachowuje spacji. Następnie potrzebujesz 4 ukośników odwrotnych.
wizzwizz4
3
Ogoliłem bajt kosztem zgodności z POSIX, tr -cd \ cdrt\\-\ Wypróbuj online!
wizzwizz4
14

C (gcc) , wynik 232 235 239 241 242, 260 407 352 249 243 249 bajtów

fn(int*f,int*i,int*n,int**t){if(f==i)if(*t=n){}if((f==i)==('+'=='*')){if((*f==',')+(*f=='\'')+(*f=='(')+(*f==')')+(*f=='*')+(*f=='+')+(*f=='=')+(*f=='\\')+(*f=='f')+(*f=='i')+(*f=='n')+(*f=='t')+(*f=='{')+(*f=='}'))if(*n++=*f){}if(fn(++f,i,n,t)){}}}

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Funkcja przenosząca wskaźnik na początek i koniec tablicy int (gdzie koniec nie zawiera pierwszego znaku), wskaźnik na bufor i wskaźnik na inny wskaźnik int. Zapisuje wynik w buforze, a wskaźnik końcowy wyniku w ostatnim wskaźniku.

pustkowie
źródło
1
W przypadku, gdy ci to pomaga, wyzwanie zostało edytowane, aby teraz można było zignorować jedną określoną wartość bajtu, na przykład dla łańcuchów zakończonych znakiem NUL. To jednak zmniejsza wynik.
Ørjan Johansen
13

Regex, wynik: 255, 2 bajty

\\

Pasuje do postaci \.

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Wcielenie ignorancji
źródło
Który to smak regularny? Według mnie pasuje to tylko do pierwszego znaku, ale nie wszystkie w danych wejściowych.
nimi
@nimi Cóż, mój link używa .NET i nie ma globalnej flagi dla wyrażeń regularnych .NET.
Wcielenie nieznajomości
12

Galaretka , wynik 254 (4 bajty)

f⁾f⁾

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⁾xy jest skrótem od “xy”(ciąg dwóch znaków).

Mamy fILTER ciąg wejściowy w dół do tylko znaki występujące w “f⁾”.

Lynn
źródło
2
Heh, miałem przygotowaną odpowiedź, f⁾⁾fale postanowiłem zadać pytanie dotyczące pobierania ciągów Unicode przed wysłaniem.
Jonathan Allan
10

Ruby- p, wynik 248 (8 unikalnych bajtów / 483 bajtów)

eval""<<11+11+11+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1+1<<11+11+11+11+1+1<<111+1+1+1+1+1<<111+1+1+1<<11+11+11<<11+11+11+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+11+1+1<<111+1+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1<<11+11+11+11+1+1+1+1+1<<11+11+11+1<<11+11+11+1+1+1+1+1+1+1+1+1+1<<11+11+11+11+11+1+1+1+1+1<<11+11+11+1+1+1+1+1+1<<11+11+11+11<<11+11+11+1+1+1+1+1+1<<11+11+11+1+1+1+1+1+1

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GB
źródło
2
To jest wspaniałe. :)
David Ljung Madison
10

Python 2 , wynik 244 245 246, 13390 1575 bajtów

Pobiera listę liczb całkowitych

exec'%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c'%(111+1,111+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1,11+11+11+11+11+11+11+11+11+11+1,111+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,111+1,111+1+1+1+1+1+1,111+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1,111+1,11+11+11+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+11+1,111+1+1+1,11+11+11+11+11+11+11+11+11+1,11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1,11+11+11+11+11+11+11+11+11+1+1,111+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1,11+11+11+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1)

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Wynika to z następującego wiersza kodu:

print[x for x in input()if x in map(ord,"exc'%+()1,")]

∼252

Ta odpowiedź jest oczywiście teoretyczna. Struktura jest

exec'<2**51 times %>c<2**50 times %>c ... %%c%c'%<112 times -~>0%<114 times -~>0% ... %<41 times -~>0%<93 times -~>0

Teoretycznie działałoby to tak samo jak:

print[x for x in input()if x in map(ord,"exc'%-~0")]

Python 3 , wynik 240 241,54 bajtów

+1 dzięki Jonathanowi Allanowi .

lambda a:[i for i in a if i in b""" ":[]abdfilmnor"""]

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ovs
źródło
Możesz pozbyć się potrzeby \w swojej wersji Py 3, używając potrójnych cytatów b''' ':[]abdfilmnor'''. W Pythonie 2 możesz zrobić pełny program w tym samym stylu dla 242: print(''.join(p for p in input() if p in ''' '().fijnoprtu'''))chociaż nie tak dobry jak exec.
Jonathan Allan
10

Wyczyść , zdobądź 247, 234 213 208 bajtów

-5 bajtów dzięki Ørjan Johansen

== =t
t['\'':t]=['\'': ==t]
t['\\':t]=['\\': ==t]
t[':':t]=[':': ==t]
t['[':t]=['[': ==t]
t[']':t]=[']': ==t]
t['t':t]=['t': ==t]
t['=':t]=['=': ==t]
t[' ':t]=[' ': ==t]
t['
':t]=['
': ==t]
t[tt:t]= ==t
t t=t

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Używa funkcji rekurencyjnej dopasowującej wzór do usuwania znaków. Nawet nie potrzebuje StdEnv.

Obrzydliwe
źródło
2
Można to bezpośrednio przetłumaczyć na funkcję Haskell z 7 unikalnymi znakami. Taki sam wynik jak moja odpowiedź, ale znacznie mniej bajtów ( wypróbuj online! ). Miły!
nimi
@nimi, których można użyć =\ zamiast ===.
Ørjan Johansen
Myślę, że chcesz, aby przełączyć ti ttna przedostatnim wierszu.
Ørjan Johansen
To oszczędza bajtów , aby zmienić nazwę ==na tale utrzymanie ==jako synonim dla niego.
Ørjan Johansen
@ ØrjanJohansen Niezły chwyt, dzięki!
Οurous
7

05AB1E , wynik: 252 253 (wykorzystano  4  3 odrębne bajty ; łącznie 8 bajtów )

„ÃJ„„JJÃ

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Wyjaśnienie:

„ÃJ        # Push the String "ÃJ"
   „„J     # Push the string "„J"
      J    # Join them together: "ÃJ„J"
       Ã   # Only keep these three characters from the (implicit) input
           # (and output the result implicitly)

Zwróć uwagę, że kolejność ÃJi „Jw 2-znakowych ciągach znaków jest ważna, ponieważ „jest również używana w słowach słownikowych , w których co dwa znaki są słowami słownikowymi (z wyjątkiem kilku wybranych znaków, takich jak litery, cyfry itp -.). Więc „JÃbyłoby „Słowo” J, a także słownikiem słowo Ã+ następny znak (w tym przypadku „), który podobno jest słowo"causing" .

Kevin Cruijssen
źródło
6
Przekreślone 4 z odstępami jest nadal przekreślone 4
Jo King
7

Perl 5, wynik 251 (5 różnych bajtów, łącznie 8058 bajtów)

s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s>><><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s>><><<s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<^>^>^^s>^><>^s<^>^>^^s>^>s>^s>^>e>^s>^>^>^s<^>^>^^s>^><>^s>^><>^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s^^^<<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s>><<><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<^>^>^^s>^><>^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s>><<><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s>><<><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s>><<><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s>><><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<eee<e^<see><>>>^<^<^>^^^^s^s^s>^<^^^^^^^^^^^>^<^^^>>eeee

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Wykorzystuje <>^espodzbiór Turinga Perla . Powyższy kod został uzyskany poprzez uruchomienie print<>=~y<<>^es><>cdrprzez mojego konwertera .

Umorusany
źródło
7

Brachylog , wynik 249,16 bajtów

{∈"{∈\\\"&}ˢ"&}ˢ

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Nie jestem do końca pewien, jak to działa z regułami, co w przypadku Brachylog ma swoją własną specjalną stronę kodową , ale wciąż mogę wyjaśnić, jak to działa:

{                Start of inline predicate.
 ∈"              Input is a member of the string literal containing:
   {∈            the opening bracket, the member-of character,
     \\\"        backslash, quote,
         &}      ampersand, the closing bracket,
           ˢ"    and the superscript s.
             &}  Unify the input to the inline predicate with the output from the inline predicate.
               ˢ Output everything from the input that made it through the inline predicate.
Niepowiązany ciąg
źródło
7

Oktawa , wynik 248 249 (50 bajtów)

Tam miał być lepszy sposób ... i nie ma! 

@(a)a('l'=='a'==all('l'=='a'==(a==('@al()''=')')))

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Filtruje wszystko oprócz @al()'=. allFunkcją jest lepsze się any, ponieważ wykorzystywane są tylko dwie odrębne bajtów. Zauważ, że any(...)jest równa not(all(not(...))). Zaczynamy od programu

@(a)a(~all(a~=('@al~()''=')'))

Zamiast ~( not) możemy również pisać 0==. Oczywiście oznaczałoby to również konieczność odfiltrowania 0. Zamiast tego należy zauważyć, że poniższe wyrażenie jest również fałszywe (zero)

'l'=='a'

Jeśli kompatybilność MATLAB nie stanowi problemu, mogę przejść do 47 bajtów

@(a)a((l='l'=='a')==all(l==(a==('@al()''=')')))
Sanchises
źródło
7

Lenguage (z stty +brkint -ignbrk), wynik 255, 3890951 bajtów

Program składa się z 3890951 bajtów NUL (co czyni ten jeden z najkrótszych programów Lenguage w historii; 3890951 bajtów jest wystarczająco mały, aby zmieścił się na moim dysku, więc uruchomiłem go w interpreterze Lenguage). OP chciał, aby rozwiązanie Lenguage / Unary zostało zagrane w golfa, więc proszę bardzo. (Zauważ, że Unary byłby znacznie dłuższy, ponieważ wymaga użycia 0zamiast NUL.)

Zauważ, że Lenguage, pomimo tego, co sugeruje jego dokumentacja, nie działa jak pieprzenie mózgu; I / O działa zupełnie inaczej (coś, co zauważyłem podczas testowania tego programu). W szczególności Lenguage odmawia pobierania danych wejściowych z niczego innego niż terminal, więc bajty, które są filtrowane, są nieprzetworzonymi bajtami wysyłanymi przez połączenie terminalowe (zwróć również uwagę, że filtrując nieprzetworzone bajty, nie zobaczysz klucze, które w ogóle wpisujesz). W praktyce oznacza to, że program będzie wchłaniał wszelkiego rodzaju dane wejściowe wysyłane do niego, z wyjątkiem bajtu NUL (zwykle wpisywanego jako Ctrl-@), który będzie dosłownie powtarzany (w tym momencie zdecydowana większość terminali go zignoruje, ponieważ bajt NUL jest końcowym odpowiednikiem instrukcji NOP). Aby sprawdzić, czy program działa, „

Co dzieje się w EOF? Cóż, jeśli terminal wysyła serię bajtów, nie ma możliwości wysłania EOF; wszystkie 256 możliwych bajtów jest interpretowanych dosłownie i nic więcej nie można wstawić do strumienia terminali. Jeśli jednak używasz starego terminala szeregowego, możesz nacisnąć przycisk „przerwa” na terminalu, aby celowo wysłać błędnie zakodowane dane, co pozwala uzyskać 257-ty możliwy kod; ta „przerwa” jest jedynym możliwym odpowiednikiem EOF, ponieważ jest wysyłany poza pasmem i wskazuje na coś innego niż ważne dane. Jeśli twoja konfiguracja terminala ma ustawioną flagę „przerwać przy zerwaniu” (a interpreter Lenguage nie zmienia, o ile wiem, tego ustawienia), wysłanie sygnału przerwania spowoduje awarię interpretera Lenguage, który dogodnie działa jako sposób na wdrożenie pożądanego zachowania EOF. Nie jestem pewien, czy jest to ustawienie domyślne (ponieważ nikt tak naprawdę nie używa terminali szeregowych, w zasadzie nigdy się nie pojawia), więc wspomniałem o tym w nagłówku jako część specyfikacji używanego interpretera języka.

Wyjaśnienie

  1110110101111100000111
001                       Initialise tape element 0 to -1
   110               111  While tape element 0 is nonzero:
      110   111              While tape element 0 is nonzero:
         101                    Read a byte from the terminal into tape element 0
               100           Output tape element 0
                  000        Add 1 to tape element 0

Wewnętrzna pętla zakończy się dopiero po wpisaniu NUL na terminalu; potem natychmiast echo wpisanego znaku (tj. NUL). Dodanie 1 w tym punkcie sprawi, że element taśmy 0 będzie ponownie niezerowy, więc zewnętrzna pętla nie będzie w ogóle mogła wyjść (dopóki wejście przerwania nie spowoduje awarii interpretera), i wrócimy do wewnętrznej pętli.

Najbardziej golfowym jest użycie odejmowania, aby wejść do zewnętrznej pętli, ale dodanie, aby kontynuować pętlę wokół niej; dodawanie ma krótsze kodowanie, ale nie może pojawić się na początku programu (ponieważ kodowanie wyglądałoby jak zera na początku i dlatego zostanie zignorowane).

ais523
źródło
Nawiasem mówiąc, jestem prawie pewien, że mogę uruchomić dowolny program Lenguage <4 TB, mimo że nie mam tak dużego dysku. Tłumacz nie dba o pliki z dziurami.
Joshua
truncatenie tworzy dla mnie rzadkich plików (może system plików ich nie obsługuje?). Jednak, o ile mogę stwierdzić, referencyjny interpreter Lenguage odczytuje plik wejściowy po linii (!), Więc jeśli program nie zawiera znaków nowego wiersza, wystarczająco długi program skończy się wyczerpaniem dostępnej pamięci. Oczywiście można to naprawić za pomocą nowego tłumacza.
ais523
Nie obcinaj. Wyszukaj nowy plik do jednego bajtu przed pożądanym końcem i wpisz bajt zerowy.
Joshua
6

Siatkówka 0.8.2 , 12 bajtów, wynik: 250

[^]n\\\n[^]

Wypróbuj online! Odfiltrowuje wszystkie znaki z wyjątkiem nowej linii, [, \, ], ^i n. Chociaż postacie [\]^są z rzędu, użycie zakresu faktycznie kosztowałoby bajt, a także obniżyłoby mój wynik.

Neil
źródło
6

Japt , 249 punktów, 20 bajtów

f"[ fq\"[\\\\\\]]" q

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Tylko ASCII
źródło
Luisowi udało się znaleźć optymalne rozwiązanie, ale możesz poprawić swój wynik za pomocą oQ+"+oQ"Qlub po prostu, ooale to tylko zmiana rozwiązania Luisa.
Kudłaty
Haha, tbh Pośpieszyłem temu i wczoraj wieczorem zabrakło mi czasu. Imo też jest krótszy i zbliża się do rozwiązania Luisa: P
tylko ASCII
4

Ruby -paF[^$_=F*'], zdobądź 250, 8 bajtów

$_=$F*''

Nie jestem wcale pewien, czy jest to zgodne z obowiązującymi zasadami witryny dotyczącymi flag wiersza polecenia, ponieważ „flaga” tutaj jest rodzajem przekazywania kodu zawierającego zabronione bajty. FFlag definiuje regex separator wejściowy składa się z wszystkich znaków w kodzie nie właściwej. Właściwy kod ustawia dane wyjściowe do konkatenacji wszystkich rekordów wejściowych.

histocrat
źródło
4

MATL , wynik 252

t't''m)'m)

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Objaśnienie (dla danych wejściowych input)

t         % Duplicate input. 
          % Stack: {'input', 'input'}
't''m)~'  % Push the string t'm)~ 
          % Stack: {'input', 'input', 't'm)~'}
m         % 'ismember' to filter
          % Stack: {'input', [0 0 0 0 1]}
)         % Index into input.
          % Stack: {'t'}
          % Implicit display.
Sanchises
źródło
3

PHP, wynik 234, 58 62 bajtów

<?=preg_replace("_[^\"$(),;<=?[\]^\_\\\\aceglnpr]_","",$argn);

Uruchom jako potok z -nFlub spróbuj online .

Tytus
źródło
3

K, wynik 246,23 bajtów

{x@&~x in"\"~ in@&{x}"}

{..} jest lambda z arg x

czytane jako „zwraca x w miejscu gdzie nie x w ciągu”

Ciąg zawiera \ "~ w @ & {x} (dziesięć bajtów używanych w programie)

J. Sendra
źródło
3

sfk , wynik 239,133 bajtów

+xed -i -case ixixi x+x+x xexex xdxdx x-x-x xixix xcxcx xaxax xsxsx xbxbx xyxyx xtxtx x"x"x x\[x\[x x\]x\]x x\\x\\x "x x x" x[byte]xx

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Wykonuje zamiany na wejściu (zapisane jako [delim][from][delim][to][delim]), zmieniając wszystkie znaki używane w programie na siebie, co usuwa pozycje znaków z rozważań na temat wprowadzania do ostatecznej zamiany, x[byte]xxktóra odwzorowuje wszystkie bajty na pusty ciąg.

Obrzydliwe
źródło
3

JavaScript, wynik 250, 95573 59273 bajtów

Lekko golfowe rozwiązanie JS, które następnie jest uruchamiane przez algorytm JSFuck . Dużo bajtów, ale dobry wynik.

Oryginał:

a=t=>t.replace(/[^![\]+()]/g,'')

Renderowanie JSFuck ma 59273 znaków:

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tschwab
źródło
2
s=>s.replace(/[^![\]+()]/g,'')powinien działać
tylko ASCII
Tak, zrobiło to znacznie. Dobry pomysł. Uważam, że bez względu na wszystko, będzie to rzędu tysięcy bajtów.
tschwab
3

pieprzenie mózgu , Wynik: 248, 162 bajtów

,[>---[----------->--<]>-[>->->->-->------->-->-------[<<]>-]>--->-->->--->------------->->-----------[<]<<[->>>->->->->->->->-<<<<<<<<<<]>>>[>]>[[>]<[->-<]>.>],]

Wypróbuj online!

Zwykle nie odpowiadam na własne pytania, ale wydaje się, że ludzie czytali moją drobną stronę, szczególnie tam, gdzie jest napisana If someone comments on a question "I wish there was a brainfuck solution", I feel obliged to provide one.

Odfiltrowane bajty są -<>[],.i ignorują bajt NUL, ponieważ nie można go odróżnić od EOF. Jest to trochę dłużej niż zwykle, ponieważ celowo unikam +uzyskiwania nieco lepszego wyniku. Mimo to może uda mi się zagrać w część generowania liczb, aby zaoszczędzić kilka bajtów.

Jo King
źródło
1
Zarówno zgłoszenia C, jak i Lenguage dostosowały się już dość konkretnie do potrzeby odróżnienia NUL od EOF. Być może korzystasz z 16-bitowego pliku BF z inną konwencją EOF?
Ørjan Johansen
3

Biała spacja , wynik: 253 (użyte 3 różne bajty, łącznie 97 bajtów)

[N
S S N
_Create_Label_LOOP][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_STDIN_as_character][T   T   T   _Retrieve][S N
S _Duplicate_input][S S S T S S T   N
_Push_9][T  S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][S N
S _Duplicate_input][S S S T S T S N
_Push_10][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][S N
S _Duplicate_input][S S S T S S S S S N
_Push_32][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT][T  N
S S _Print_character][N
S N
N
_Jump_to_Label_LOOP]

Litery S(spacja), T(tab) i N(nowa linia) dodane tylko jako wyróżnienia.
[..._some_action]dodano tylko jako wyjaśnienie.

Wypróbuj online (tylko z surowymi spacjami, tabulatorami i nowymi wierszami). Fajną zaletą jest to, że mogę użyć tego kodu do przekształcenia podświetlonego kodu powyżej w surowy program Whitespace (co zrobiłem w tym TIO jako przykład), do którego zwykle używam Notepad ++. :)

Wyjaśnienie:

Całkiem prosto. Oto pseudo-kod:

Loop indefinitely:
  Character c = STDIN as character
  if(c == SPACE or c == TAB or c == NEWLINE):
    Print c as character
  Go to the next iteration of the loop
Kevin Cruijssen
źródło