Evaluating Parentheses and Brackets as Integers

CJam, 23

q"])(""1]:*0]:+["4/ers~

With BIG credits to Dennis! Try it online

Explanation:

The program converts the input to a CJam expression then evaluates it.
[…] becomes […1]:* (append 1 and multiply)
(…) becomes […0]:+ (append 0 and add)

q              read input
"])("          characters we want to replace
"1]:*0]:+["    replacement strings, concatenated
4/             split into strings of length 4: ["1]:*" "0]:+" "["]
er             replace (transliterate) the 3 characters with the 3 strings
s              convert the result (mixed characters and strings) to string
~              evaluate

Common Lisp - 98

(lambda(s)(eval(read-from-string(#1=ppcre:regex-replace-all"\\["(#1#"]"(#1#"\\("s"(+")")")"(*"))))

1. Replace ( by (+
2. Replace [ by (*
3. Replace ] by )
4. Read from string
5. Eval

This requires cl-ppcre library to be loaded in current lisp image.

Explanation

Functions * and + are variadic and return their neutral value when given no arguments. For your examples, the evaluated lisp form are the following ones:

(+ (*) (+ (+)) (+ (*) (*)) (* (+)) (* (+ (* (*) (*)) (*) (*)) (+ (*) (*))))
=> 9

and

(* (+ (*) (*) (*) (*) (*)) (+ (*) (*) (*)) (+ (*) (*) (*))
   (+ (+ (+ (+ (+ (+ (*) (*))))))))
=> 90

Without regexes - 183 bytes

(lambda(s)(do(r(x(coerce s'list))c)((not x)(eval(read-from-string(coerce(reverse r)'string))))(setq c(pop x))(push(case c(#\[ (push #\* r)#\()(#\] #\))(#\( (push #\+ r) #\()(t c))r)))

C'mon, Lisp - 16 bytes (experimental)

+((<r*([<r)]<rRE

Other languages are so terse that I am tempted to make my own golfing language based on Common Lisp, for shorter string manipulations. Currently there is no spec, and the eval function is the following one:

(defun cmon-lisp (expr &rest args)
  (apply
   (lambda (s)
     (let (p q)
       (loop for c across expr
             do (case c
                  (#\< (push (pop p) q))
                  (#\r
                   (let ((a1 (coerce q 'string)) (a2 (coerce p 'string)))
                     (setf p nil
                           q nil
                           s
                             (cl-ppcre:regex-replace-all
                              (cl-ppcre:quote-meta-chars a1) s a2))))
                  (#\R
                   (setf s
                           (if (string= s "")
                               nil
                               (read-from-string s))))
                  (#\E (setf s (eval s)))
                  (t (push c p))))
       s))
   args))

Tests:

(cmon-lisp "+((<r*([<r)]<rRE" "([] [] ([] []))")
=> 4

• there is an implicit argument called s and two stacks, p and q.
• characters in source code are pushed to p.
• < : pops from p and pushes to q.
• r : replaces in s (must be a string) from characters in q to charactes in p; result is stored in s; p and q are emptied.
• R : read from string s, store result in variable s.
• E : eval form s, store result in s.

Pyth, 35 34 33 bytes

L?*F+1yMbqb+YbsyMbyvsXzJ"])"+R\,J

Demonstration.

1 byte thanks to @Jakube.

We start by parsing the input. The input format is close to Python, but not quite. We need commas after each parenthesized or bracketed group. The comma at the end of a bracketed group is unnecessary, but harmless. To accomplish this, we use this code:

vsXzJ"])"+R\,J
  X               Translate
   z              in the input
     "])"         the characters "])"
    J             which we will save to J to
             J    J
         +R\,     with each character mapped to itself plus a ",".
 s                Combine the list to a string.
v                  Evaluate as a Python literal.

This will leave an extra , at the end of the string, which will wrap the whole object in a tuple, but this is harmless, because the tuple will be summed, and so have a value equal to its element.

Now that the string is parsed, we must find its value. This is done using a user defined function, y, which is called on the parsed object. the function is defined as follows:

L?*F+1yMbqb+YbsyMb
L                     Define a function, y(b), which returns the following:
 ?       qb+Yb        We form a ternary whose condition is whether the input, b,
                      equals the inputplus the empty list, Y. This is true if
                      and only if b is a list.
      yMb             If so, we start by mapping y over every element of b.
  *F+1                We then take the product of these values. The +1 ensures
                      that the empty list will return 1.
                yMb   Otherwise, we start by mapping y over every element of b.
               s      Then, we sum the results.