是否可以提供自定义范围，将remove vars添加到eval函数？

Question

I'm trying to use eval() to evalute a mathematical string with variables and functions

ex: algo = "1+len+customfunction(6)"

So i have data for len and the function for customFunction .

They are obviously declared in different scope.

I tried with something like

process = function(vars, algo) {
 return (function() {
    algo = algo.toLowerCase();
    return eval(algo);
  }).call(vars);
};

I need to provide required functions and variables to eval . Items are in different scopes, how do i do that ?

Now I'm a bit lost and confused, is this even possible ? I think using eval('var'+vName+'='+value) would be ok for vars but not suitable for functions.

EDIT : btw eval can be replaced with (new Function(algo))() http://moduscreate.com/javascript-performance-tips-tricks/

Answer 1

i would move the required data to an object, which hold all required items. access is possible with any given string. this solution does not requires eval()

var data = {
    customfunction: function (x) {
        return Math.PI * x * x;
    },
    len: 5
};
var variable = 'len';
if (data[variable]) {
    // do something
    ;
}
var fn = 'customFunction';
function evaluate(vars, algo) {
    if (data[algo.toLowerCase()]) { // 
        return data[algo.toLowerCase()].call(vars);
    } else {
        // fallback
    }
}
var process = evaluate(vars, algo);

Answer 2

OK here is what i've found :

• we can't provide a context for eval
• eval('1+1') can be replaced by ((new Function('1+1'))() and it's way faster
• eval have different scopes, can't use eval('var a = 1'); eval('alert(a)');

So, i'v managed to :

• create a string to initiate all vars : vars += 'var '+key+'='+JSON.stringify(value)+';'

• replace at runtime my algo string functions with their full path :

  var algo = 'f(x)+5*2', x = 5;

  window.object.functions.x = function(a) { return a*2; };

  So at runtime i replace x( with object.functions.x( the code executed by eval finally is eval('object.functions.x(5)+5*2'); and output is 20 of course :)

It seems to work.