Write a brain-flak classic interpreter!

Question

Brain-Flak (a cross between Brainf**k and Flak-Overstow) is a stack-based esoteric language. Since this challenge was posted, the language has evolved and updated, but this first revision of the language is known as "brain-flak classic".

You must write a program or function that takes a string of Brain-Flak classic code, and evaluates it. It will also take a (possible empty) list of integers. There are the inputs to the Brain-Flak classic program.

The Language

Brain-Flak has two stacks, known as 'left' and 'right'. The active stack starts at left. If an empty stack is popped or peeked, it will return 0. There are no variables. When the program starts, each input is pushed on to the active stack in order (so that the last input is on top of the stack).

The only valid characters in a Brain-Flak program are ()[]{}<>, and they must always be balanced. If there are invalid characters, or the brackets are unmatched, you get undefined behaviour. Anything is valid.

There are two types of functions: Nilads and Monads. A nilad is a function that takes 0 arguments. Here are all of the nilads:

• () +1.
• [] -1.
• {} Pop the active stack.
• <> Toggle the active stack.

These are concatenated together when they are evaluated. So if we had a '3' on top of the active stack, this snippet:

()(){}

would evaluate to 1 + 1 + active.pop() which would evaluate to 5. <> evaluates to 0.

The monads take one argument, a chunk of Brain-Flak code. Here are all of the monads:

• (n) Push 'n' on the active stack.
• [n] Print 'n' as an int and a newline.
• {foo} While active.peek() != 0, do foo. Evaluates to 0¹.
• <foo> Execute foo, but evaluate it as 0.

These functions will also return the value inside of them, so

(()()())

Will push 3 and

[()()()]

Will print 3 but

[(()()())]

Will print and push 3.

When the program is done executing, each value left on the active stack is printed as an integer, with a newline between. Values on the other stack are ignored.

Rules:

• Your program must support numbers in the (-128, 127) range, and a stack size of at least 255. If you support larger, great.

• Underflow/overflow is undefined.

Sample IO:

The empty program:

Input: None

Output: None

Addition. Source:

({}{})

Input:

2, 3

Output:

5

Subtraction. Source:

({}<>){({}[])<>({}[])<>}<>

Input:

2, 3

Output:

-1

Multiplication. Source:

({}<>)<>({}[]){({}[])<>(({}))<>}<>{({}<>{})<>}<>

Input:

7, 8

Output:

56

Fibonacci. Source:

<>((()))<>{({}[])<>({}<>)<>(({})<>({}<>))<>}<>

Input:

5

Output:

13
8
5
3
2
1
1

Truth machine

{[({})]}

Standard loopholes apply, and the shortest answer in bytes wins.

---

• ¹: This was actually a mistake on my part. {...} should evaluate to the sum of all of it's runs, which is IMO one of the coolest features of brain-flak. However, for the purposes of this challenge, assume that {...} evaluates as 0.

Answer 1

Pip -n, 151 148 101 98 bytes

YRVg;VqR^"{}()<>[]";,8R J,8<>2AL,8("POy|i o0Syl1v0W@y{ }1yPU$+[ ]&@y0 1P$+[ ]"R0" (V{"R1"i}) "^s)y

Takes the list of inputs as command-line arguments and the Brain-Flak code from (a line of) stdin. Try it online!

Edit: Saved a whole lot of bytes over my original approach by switching to a translate-and-eval strategy.

Ungolfed and commented

This version also includes some debug output showing the Pip code that results from the translation, as well as the stack contents after execution.

;;; Setup ;;;

; y is the active stack, l is the off-stack
; y is initialized from command-line arguments
y:RVg   (reversed to put the last input at the top)
; l is preset to empty list by default

; p is the program (read from stdin)
p:q

; Translate from braces to numbers 0-7 (we do this so that the
; later replacement step won't try to replace the braces in the
; Pip code)
p R: ^"()[]{}<>" 0,8

;;; Replace nilads with the appropriate code ;;;

; () => o (variable preset to 1)
p R: 01 "o"

; [] => v (variable preset to -1)
p R: 23 "v"

; {} => POy|i
; Pop y; return that value OR i (variable preset to 0)
p R: 45 "POy|i"

; <> => (V{Syli})
; Eval the code Syl to swap stacks y and l, then return i (i.e. 0)
p R: 67 "(V{Syli})"

;;; Replace monads with the appropriate code ;;;

; ( ) => yPU$+[ ]&@y
; Sum ($+) the inside and push (PU) the sum onto y; return
; the just-pushed value, which is the first element of y (@y)
; y will always be truthy (nonempty), since we just pushed a value onto it
p R: 0 "yPU$+["
p R: 1 "]&@y"

; [ ] => P$+[ ]
; Sum ($+) the inside, print (P) the sum, and return it
p R: 2 "P$+["
p R: 3 "]"

; { } => (V{W@y{ }i})
; Eval the code W@y{ }, which wraps the inside in curly braces
; and runs it while (W) the first element of y (@y) is truthy
; (i.e. not zero, and not nil from an empty stack)
; Then return i (i.e. 0)
p R: 4 "(V{W@y{"
p R: 5 "}i})"

; < > => (V{ i})
; Eval the inside, then return i (i.e. 0)
p R: 6 "(V{"
p R: 7 "i})"

; Debug: print the resulting translated code and a blank line
Pp.n

;;; Run the code ;;;

; Eval the translated code
(Vp)

; Output the active stack, newline-separated
PyJn

; Debug: print the active stack and the off-stack
P"Active stack: ".RPy
"Off-stack: ".RPl

Answer 2

Brain-Flak Classic, 1271 1247 1239 bytes

<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}

Try it online!

+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.

<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>{}){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}

Try it online!

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
{({}[]<{}>)}{}

Answer 3

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.

Answer 4

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←⍬

Try it online!

one Classic interpreting another :)

Answer 5

Python 3, 429 bytes

import re
S='s+=[v];v=0';T='v+=s.pop()';i=0
d={'()':'v+=1','(':S,')':'a+=[v];'+T,'[]':'v-=1','[':S,']':'print(v);'+T,'<>':'a.reverse()','<':S,'>':T,'{}':'v+=0if a[-1]==""else a.pop()','{':S+';while a[-1]:','}':T}
def r(m):global i;t=m.group();i-=(t=='}');s=' '*i;i+=(t=='{');return''.join(s+r+'\n'for r in d[t].split(';'))
def g(c,*a):
 a,s,v=['']+list(a),[],0;exec(re.sub(r'[<({[]?[]})>]?',r,c));
 while a[-1]!="":print(a.pop())

Used like g('[{}{}]', 2, 3)

It uses re.sub to "compile" the brain-flak source to python and then executes the python. (for debuging, replace exec with print to get a listing of the python code)

Properly indenting nested while loops eats up a lot of bytes in the code.

Answer 6

Perl 5.6, 419 414 bytes

I've golfed it a bit but there's probably scope for improvement. Newlines and tabs added here for the sake of a bit of readability:

use Text::Balanced extract_bracketed;
$s=shift;
@a=reverse@ARGV;
sub p
{
    my($c)=@_;
    my$s=0;
    while(my$n=extract_bracketed($c)){
        $s+='()'eq$n||'{}'eq$n&&shift@a;
        $s-='[]'eq$n;
        @t=@a,@a=@i,@i=@t if'<>'eq$n;
        my$m=chop($n);
        $n=substr($n,1);
        if($n){
            p($n)while'}'eq$m&&$a[0];
            p($n)if'}'ne$m;
            $s+=$v,unshift@a,$v if')'eq$m;
            $s+=$v,print"n=$n m=$m v=$v\n"if']'eq$m;
        }
    }
    $v=$s;
}
p($s);
foreach(@a){
    print"$_\n";
}

Answer 7

Python, 616 bytes

Instructions:

1. Run with python
2. Input list in [1,2,...] format, then press enter
3. Paste/write program, then press enter again
4. Done

Basically, what this program does is recursively "compile" the Brain-flak code into nested lists, and recursively interpret that list. There probably is a way to combine the two...

I'll try and rework the logic later.

y="([{<)]}>"
w,z,g=print,len,input
def c(s):
 if z(s)<1:return[]
 t,i,o=[],1,0
 t.append(y.index(s[0]))
 while z(t)>0:
  x=y.index(s[i])
  if x<4:t.append(x)
  else:o=t.pop()
  i+=1
 r=[[o,c(s[1:i-1])]]
 r.extend(c(s[i:]))
 return r
p=lambda t:t.pop()if z(t)>0 else 0
k=lambda t:t[z(t)-1]if z(t)>0 else 0
r,l=[],eval(g())
a=l
def i(u):
 v=0
 global a
 for t,n in u:
  if t<1:
   if n:o=i(n);v+=o;a.append(o)
   else:v+=1
  if t==1:
   if n:o=i(n);v+=o;w(o)
   else:v-=1
  if t==2:
   if n:
    while k(a)!=0:i(n)
   else:v+=p(a)
  if t>2:
   if n:i(n)
   elif a==l:a=r
   else:a=l
 return v
i(c(g()))
for n in a:w(n)

Answer 8

Python 2, 361, 348 bytes

c,s=input();s=s,[]
a=s[0]
def w():global a,s;s=s[::-1];a=s[0];return 0
def p(c):a.append(c);return c
def n(c):print c;return c
z=lambda c:0
def l(f):
 global a
 while a and a[-1]:f()
 return 0
for x,y in zip("() ( [] {} <> [ < { } ] >".split(),"+1 +p( -1 +(len(a)and(a.pop())) +w() +n( +z( +l(lambda: ) ) )".split()):c=c.replace(x,y)
exec c
print a

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