How to call and execute an operator from string?

Question

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()

Looks like one of the few cases where eval() would be handy. — alex
– alex, Commented Sep 9, 2011 at 1:03
You might find the following series interesting: Essentials of Interpretation. They are small lesson about computer program interpretation, written in Javascript, the goal at the end IMO will be to implement a small scheme-like language. — Christian C. Salvadó
– Christian C. Salvadó, Commented Sep 9, 2011 at 1:20

Community · Accepted Answer · 2017-05-23 10:29:14Z

7

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));

edited May 23, 2017 at 10:29

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answered Sep 9, 2011 at 1:04

Joe

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1 Comment

jfriend00 Over a year ago

+1 for using the object to look up operator functions instead of if/else or switch/case.

Mike Samuel · Accepted Answer · 2011-09-09 01:04:51Z

1

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;
  }
}

answered Sep 9, 2011 at 1:04

Mike Samuel

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Comments

maerics · Accepted Answer · 2011-09-09 01:07:03Z

0

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);
});

answered Sep 9, 2011 at 1:07

maerics

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