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    <h2>Ruby</h2> <p><strong>Number of characters: <del>217</del> 179</strong></p> <p><del>This is the shortest ruby solution up to now (one heavily based on RegExp yields incorrect answers when string contains few groups of parenthesis)</del> -- no longer true. Solutions based on regex and substitution are shorter. This one is based on stack of accumulators and parses whole expression from left to right. It is re-entrant, and does not modify input string. It could be accused of breaking the rules of not using <code>eval</code>, as it calls <code>Float</code>'s methods with identical names as their mathematical mnemonics (+,-,/,*).</p> <p>Obfuscated code <em>(old version, tweaked below)</em>:</p> <pre><code>def f(p);a,o=[0],['+'] p.sub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n| q,w=n;case w;when'(';a&lt;&lt;0;o&lt;&lt;'+';when')';q=a.pop;else;o&lt;&lt;w end if q.nil?;a[-1]=a[-1].method(o.pop).call(q.to_f) if !q.nil?};a[0];end </code></pre> <p>More obfuscated code:</p> <pre><code>def f(p);a,o=[0],[:+] p.scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|q,w=n;case w when'(';a&lt;&lt;0;o&lt;&lt;:+;when')';q=a.pop;else;o&lt;&lt;w;end if !q a&lt;&lt;a.pop.send(o.pop,q.to_f)if q};a[0];end </code></pre> <p>Clean code:</p> <pre><code>def f(p) accumulators, operands = [0], ['+'] p.gsub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each do |n| number, operand = n case operand when '(' accumulators &lt;&lt; 0 operands &lt;&lt; '+' when ')' number = accumulators.pop operands.pop else operands[-1] = operand end if number.nil? accumulators[-1] = accumulators.last.method(operands[-1]).call(number.to_f) unless number.nil? end accumulators.first end </code></pre>
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    1. POEvaluating a string of simple mathematical expressions
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    1. COActually, mine is shorter (198) and uses regex to solve parenthesis from the top down before the final result of the mathematics. So "3 + (3 * (3 + 9))" goes: "3 + (3 * 12)", "3 + 36", 39. It goes top-to-bottom, left-to-right. It solves all tests, except one minor oversight which requires spaces between tokens. See: http://stackoverflow.com/questions/928563/code-golf-evaluating-mathematical-expressions/932627#932627
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    2. CONot that yours isn't clever, it very much is.
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    3. CO(3+7) - (5+2) -- that's what I've meant by several parentheses groups. Your solution has problem with non-nested parentheses, because of regex greediness.
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