for example:

var s = '3+3';
s.replace(/([\d.]+)([\+\-)([^,]*)/g,
        function(all, n1, operator, n2) {
                r = new Number(n1) ??? new Number(n2);
                return r;
        }
);

note: not using eval()

for example:

var s = '3+3';
s.replace(/([\d.]+)([\+\-)([^,]*)/g,
        function(all, n1, operator, n2) {
                r = new Number(n1) ??? new Number(n2);
                return r;
        }
);

note: not using eval()

• javascript
• operators

• Looks like one of the few cases where eval() would be handy. – alex Commented Sep 9, 2011 at 1:03
• Is new Function() off limits too? – alex Commented Sep 9, 2011 at 1:07
• Why not using eval() out of curiosity? – Demian Brecht Commented Sep 9, 2011 at 1:10
• You might find the following series interesting: Essentials of Interpretation. They are small lesson about puter program interpretation, written in Javascript, the goal at the end IMO will be to implement a small scheme-like language. – Christian C. Salvadó Commented Sep 9, 2011 at 1:20

3 Answers 3

Are Variable Operators Possible?

Not possible out of the box, but he gives a nice implementation to do it, as follows. Code by delnan.

var operators = {
    '+': function(a, b) { return a + b },
    '<': function(a, b) { return a < b },
     // ...
};

var op = '+';
alert(operators[op](10, 20));

So for your implementation

r = operators[operator](new Number(n1), new Number(n2));

Your regex is a bit broken.

/([\d.]+)([\+\-)([^,]*)/g

should probably be

/([\d.]+)([+-])([\d+]+)/g

then you can switch on the operator:

function (_, a, op, b) {
  switch (op) {
    case '+': return a - -b;
    case '-': return a - b;
  }
}

s.replace(/(\d+)\s*([+-])\s*(\d+)/g, function(all, s1, op, s2) {
  var n1 = Number(s1), n2 = Number(s2);
  return (op=='+') ? (n1+n2) : (n1-n2);
});