Evaluating a string of simple mathematical expressions [closed]

Answer 1

SNOBOL4

Number of characters: 232

        a = pos(0) | '('
        n = span('0123456789.')
        j = '!+;!-;!*;!/;       output = e'
d       j '!' len(1) . y = "    e a . q n . l '" y "' n . r = q (l " y " r)     :s(p)"  :s(d)
        k = code(j)
        e = input
s       e ' ' = :s(s)
p       e ('(' n . i ')') = i   :s(p)f
end

This is a semi-cheat. It uses code() (a variant of eval) to de-compress itself, but not to evaluate the input expression.

De-obfuscated version, without code:

        prefix = pos(0) | '('
        num = span('0123456789.')
        expr = input
spaces  expr ' ' = ''   :s(spaces)
paren   expr ('(' num . x ')') = x      :s(paren)
add     expr (prefix . pfx) (num . l) '+' (num . r) = pfx (l + r)       :s(paren)
sub     expr (prefix . pfx) (num . l) '-' (num . r) = pfx (l - r)       :s(paren)
mul     expr (prefix . pfx) (num . l) '*' (num . r) = pfx (l * r)       :s(paren)
div     expr (prefix . pfx) (num . l) '/' (num . r) = pfx (l / r)       :s(paren)
        output = expr
end

Strategy:

• First, remove all spaces (spaces)
• Whenever possible, remove parentheses surrounding a number (paren)
• Otherwise, find a simple expression involving two numbers, prefixed by '(' or at the start of the string
• If none of the above rules apply, the expression is fully evaluated. Now if the input was well formed we should be left with a number.

Example:

• 1 + (2 * 3) + 4
• 1+(2*3)+4 [spaces]
• 1+(6)+4 [mul]
• 1+6+4 [paren]
• 7+4 [add]
• 11 [add]