Write a brain-flak classic interpreter!
Brain-Flak Classic, 1271 1247 1239 bytes
<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}
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+4 bytes from fixing a bug with the condition in the {...} monad, and -36 bytes from various golfs.
1238 bytes of code, +1 byte for the -a flag (which can be combined with the language flag).
This now evaluates {...} as zero per the challenge specification. Note that Brain-Flak itself has evaluated {...} as the sum of all runs since the May 7, 2016 bugfix two days before this challenge was posted.
The following code interprets Brain-Flak Classic correctly, with {...} as the sum of all runs. The only difference between the two interpreters is the placement of one {} nilad.
<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>{}){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}
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Input (to either interpreter) is the Brain-Flak Classic program to interpret, then a newline, then a space-separated list of integers. No validation is performed on the input. The newline is required, even if the program or input is blank.
The first step is to parse all of the input, starting with the brackets:
# Move to right stack, and push 1 to allow loop to start
<>(())
{
# While keeping -5 on third stack:
<>((([][][][][])<
# Pop bracket or newline k from left stack, and push 0, k-10, k-40, k-60, k-91, k-123 on right stack
(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])
# Search this list for a zero, and push the number of nonzero entries popped minus 5
# (thus replacing the 0 if it was destroyed)
>{()<{}>}{})
# Remove rest of list, and push the same number plus 1
# Result is -4 for {, -3 for [, -2 for <, -1 for (, 0 for newline, or 1 for everything else (assumed closing bracket)
<{{}}{}>())
# Repeat until newline found
}{}<>
Then the integers are parsed. This wouldn't normally be required, but the input was taken as ASCII. This does have a silver lining, though: text input allows us to determine the stack height, which simplifies things when we don't have access to the stack height nilad.
Integers are parsed into two numbers on the second stack: one for the absolute value, and one for the sign. These are then moved back to the first stack.
The interpreted stacks are stored below the code on the first stack in the following order: current stack height, current stack, other stack height, other stack. The 0 for the other stack height does not need to be pushed at this point, since it will be an implicit zero the first time it is read.
(<((
# If stack nonempty, register first stack entry.
{()(((<>))<>)}{}
# For each byte k of input:
{
# Push -3, -13, and k-32
<({}(([][][])((({})({}))[]{})){})>
# Evaluate to 1 if space
# If not space (32):
((){[]<
# If not minus (45):
({}{})((){[]<
# Replace top of right stack (n) with 10*n + (k-48)
({}{}<>((({})({})){}{}){})(<>)
# Else (i.e., if minus):
>}{}){
# Remove excess "else" entry and -3
{}{}
# Set sign to negative (and destroy magnitude that shouldn't even be there yet)
<>(<({}{}())>)(<>)}
# Else (i.e., if space):
>}{}){
# Remove working data for byte, and push two more 0s onto right stack
(<{}{}{}((<>))<>>)
# Push number of integers found
}{}}<>)
# For each integer:
<{({}[]<
# Move magnitude back to left stack
({}<>)<>
# If sign is negative, negate
{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}
>)}{}
# Push stack height onto stack
<>>)
# Push 0
>)
The representation of the code is now moved back to the left stack. To make things easier later, we subtract 4 from the opening brackets of nilads, so that each operation has a unique integer from -1 to -8.
# For each bracket in the code:
<>{
# Push k-1 and evaluate to k
(({}[])()
# If not closing bracket:
{
# Check next bracket (previously checked, since we started at the end here)
(<{}>)<><(({})[])>
# Subtract 4 if next bracket is closing bracket
# Inverting this condition would save 8 bytes here, but cost 12 bytes later.
[][][][]{()()()()(<{}>)}{}
<>}{}
# Push result onto left stack
<>)
<>}<>{}
The main part of the program is actually interpreting the instructions. At the start of each iteration of the main loop, the current instruction is at the top of the left stack, everything after it is below it on the same stack, and everything before it is on the right stack. I tend to visualize this as having a book open to a certain page.
{
(
# Get current instruction
({})
# Move all code to left stack, and track the current position in code
<({()<<>({}<>)>}{})>
# Push -1, signifying that the code will move forward to just before a matching }.
# In most cases, this will become 0 (do nothing special) before it is acted upon
([])
# Push instruction minus 1
)
# If opening bracket:
((){[](<
# Push instruction+1 and instruction+4
(({}()()(<>))()()())
# If instruction+4 is nonzero (not loop monad), replace the earlier -1 with 0 to cancel forward seek
# This would be clearer as {(<{}>)<>(<{}>)<>}, but that would be unnecessarily verbose
{(<{}>)<>}
# Else (i.e., if closing bracket):
>)}{}<>){
# If closing bracket, parse command
# Post-condition for all: if not moving to {, pop two and push evaluation, 0.
# (For nilads, can assume second from top is 0.)
# If moving to {, pop one, push -3, 0, 0.
# Seven nested if/else statements, corresponding to eight possible instruction.
# The "else" statements end with 0 already on the stack, so no need to push a 0 except in the innermost if.
# Each one beyond the first increments the instruction by 1 to compare the result with 0
# Each instruction will pop the instruction, leaving only its evaluation (with a 0 on top).
{}((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[](<
# -7: pop
# Pop instruction to reveal existing 0 evaluation
{}
# Move code out of the way to access stack
<>{({}<>)<>}{}
# Duplicate stack height (only useful if stack height is zero)
(({}))
(
# If stack height nonzero
{
# Save stack height on second stack
<{}({}<>)<>>
# Pop stack
{}
# Move stack height back and subtract 1
(<<>({}[]<>)>)
}
# Move code back to normal position
<><{({}<>)<>}>{}
# Evaluate as popped entry (0 if nothing popped)
)
# (else)
>)}{}){
# -6: -1 nilad
# Just evaluate as -1
{}{}(<([])>)
# (else)
}>}{}){
# -5: swap nilad
# Move code out of the way to access stack
{}<>{({}<>)<>}{}
# Number of integers to move: stack height + 1 (namely, the stack height and every entry in the stack)
((({})())
# Move to second stack
<{({}[]<({}<>)<>>)}>{}
# Do (stack height + 1) times again
){({}[]<><
# Get stack element
({}<><
# Move alternate (interpreted) stack to second (real) stack, and push length on top of it
({()<({}[]<({}<>)<>>)>}{}<>)
# Push current stack element below alternate stack
><>)
# Move alternate stack back above newly pushed element
<>({()<({}[]<({}<>)<>>)>}{}<>)
>)}
# Move code back to normal position
<>(<{({}<>)<>}>)
# (else)
}>}{}){
# -4: 1
# Just evaluate to 1
{}{}(<(())>)
# (else)
}>}{}){
# -3: loop
# Create zero on stack while keeping existing evaluation
# This becomes (<{}{}>) in the version that meets the challenge spec
(<{}>)
# Move code out of the way to access stack
<>{({}<>)<>}{}
# Duplicate stack height
(({}))
(
# If stack height nonzero
{
# Save stack height on second stack
<{}({}<>)<>>
# Peek at top of stack
({})
# Move stack height back
(<<>({}<>)>)
}
# Move code back to normal position
<><{({}<>)<>}>
# Look at peeked entry
# Remove the {} in the version meeting the challenge spec
{})
# If peeked entry is nonzero
{
# Replace -3 instruction on third stack
{}([][][])
# Replace loop indicator to 0 (to be incremented later to 1)
<>(((<{}>)
# Create dummy third stack entry to pop
<>))
}
# (else)
}>}{}){
# -2: print
# Just print evaluation without modifying it
{}(<([{}])>)
# (else)
}>}{}){
# -1: evaluate as zero
# Just change evaluation to 0
{}((<{}>))
# else
}>}{}){
# 0: push
# Get current evaluation (without modifying it)
{}(({})
# Create zero on stack as barrier
(<()>)
# Move code out of the way to access stack
<<>{({}<>)<>}{}
# Increment stack height and save on other stack
({}()<>)<>
# Push evaluation
>)
# Move stack height back (and push zero)
<>(<({}<>)>)
# Move code back to normal position
<>{({}<>)<>}
}{}
# Update third stack by adding evaluation to previous entry's evaluation
# Previous entry's instruction is saved temporarily on left stack
(<({}<({}<>)<>>{})<>({}<>)>)
# Increment loop indicator
# If instruction was loop monad and top of stack was nonzero, this increments 0 to 1 (search backward)
# Otherwise, this increments -1 to 0 (do nothing)
<>(<({}())>)
}{}
# While holding onto loop indicator
({}<
# Go to immediately after executed symbol
{({}[]<({}<>)<>>)}{}
>)
# If looping behavior:
{
# Switch stack and check if searching forward
((({}[]<>)
# If so:
{
# Move just-executed { back to left stack, and move with it
(<{}({}<>)>)
}{}
# Either way, we are currently looking at the just-executed bracket.
# In addition, the position we wish to move to is on the current stack.
# Push unmodified loop indicator as initial value in search
())
# While value is nonzero:
<{
# Add 1
({}()
# Move current instruction to other stack
<({}<>)<>
# Check whether next instruction is closing bracket
(({})[])>
# If opening bracket, subtract 2 from value
{[][](<{}>)}{}
)
}{}>
# If searching backward, move back to left stack
()){{}(<>)}
}{}
}
After exiting the main loop, all of the code is on the right stack. The only things on the left stack are a zero and the two interpreted stacks. Producing the correct output is a simple matter.
# Pop the zero
{}
# Output current stack
{({}[]<[{}]>)}{}
# Discard other stack to avoid implicit printing
{({}[]<{}>)}{}
APL, 255 257 bytes
b←{S←(⌽⍺)⍬
e←{0=⍴⍵:0
v+∇⊃_ v←∇{r←⊂2↓⍵
'()'≡n←2↑⍵:r,1
'[]'≡n:r,¯1
'{}'≡n:r,{i←⊃⊃⊃S⋄S[1]↓⍨←1⋄i}⍬
'<>'≡n:r,0⊣S⌽⍨←1
r←⊂⍵↓⍨i←0⍳⍨(+\c=⍵)-+\')]>}'['([<{'⍳c←⊃⍵]=⍵
i←1↓¯1↓c←i↑⍵
'('=c←⊃c:r,S[1],⍨←⍺⍺i
'['=c:r,+⎕←⍺⍺i
'{'=c:r,{0≠⊃⊃⊃S:∇e i⋄0}⍬
'<'=c:r,0⊣⍺⍺i}⍵}
⎕←⍪⊃S⊣e⍵}
This takes the program as its right argument, and the program input as its left argument, i.e:
2 3 b '({}{})'
5
2 3 b '({}<>){({}[])<>({}[])<>}<>'
¯1
7 8 b '({}<>)<>({}[]){({}[])<>(({}))<>}<>{({}<>{})<>}<>'
56
5 b '<>((()))<>{({}[])<>({}<>)<>(({})<>({}<>))<>}<>'
13
8
5
3
2
1
1
Ungolfed version: here.
APL (Dyalog Classic), 146 bytes
↑⍕¨s⊣{⍎⍕1 ¯1'(s↓⍨←1)⊢⊃s' '0⊣s t←t s' 's,⍨←+/∇¨a' '⎕←+/∇¨a' '∇{×⊃s:∇⍺⍺¨a⋄0}0' '0⊣+/∇¨a'[(⊃⍵)+4×⍬≢a←1↓⍵]}¨⍎∊(')',⍨'(',¨⍕¨⍳4)[0,4,⍨'([{<'⍳⍞]⊣s←⌽⎕⊣t←⍬
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one Classic interpreting another :)