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IIFE a clase en javascript

Tengo un pequeño problema con los módulos / IIFE, etc. Tengo un script que solía ser un IIFE y usa mucho esta palabra clave, etc. Estoy tratando de convertirlo en un módulo.

 (function () { var _symbol = d3.svg.symbol(), _line = d3.svg.line(); d3.superformula = function () { var type = _symbol.type(), size = _symbol.size(), segments = size, params = {}; function superformula(d, i) { var n, p = _superformulaTypes[type.call(this, d, i)]; for (n in params) p[n] = params[n].call(this, d, i); return _superformulaPath( p, segments.call(this, d, i), Math.sqrt(size.call(this, d, i)) ); } superformula.type = function (x) { if (!arguments.length) return type; type = d3.functor(x); return superformula; }; // size of superformula in square pixels superformula.size = function (x) { if (!arguments.length) return size; size = d3.functor(x); return superformula; }; // number of discrete line segments superformula.segments = function (x) { if (!arguments.length) return segments; segments = d3.functor(x); return superformula; }; return superformula; }; function _superformulaPath(params, n, diameter) { var i = -1, dt = (2 * Math.PI) / n, t, r = 0, x, y, points = []; while (++i < n) { t = (params.m * (i * dt - Math.PI)) / 4; t = Math.pow( Math.abs( Math.pow(Math.abs(Math.cos(t) / params.a), params.n2) + Math.pow(Math.abs(Math.sin(t) / params.b), params.n3) ), -1 / params.n1 ); if (t > r) r = t; points.push(t); } r = (diameter * Math.SQRT1_2) / r; i = -1; while (++i < n) { x = (t = points[i] * r) * Math.cos(i * dt); y = t * Math.sin(i * dt); points[i] = [Math.abs(x) < 1e-6 ? 0 : x, Math.abs(y) < 1e-6 ? 0 : y]; } return _line(points) + "Z"; } var _superformulaTypes = { asterisk: { m: 12, n1: 0.3, n2: 0, n3: 10, a: 1, b: 1 }, bean: { m: 2, n1: 1, n2: 4, n3: 8, a: 1, b: 1 }, }; d3.superformulaTypes = d3.keys(_superformulaTypes); })();

Me estoy confundiendo con la forma de definir el constructor si uso la clase, o es6 también funcionaría si alguien pudiera ayudarme. Pero no estoy seguro de cómo hacerlo. Si alguien puede ayudar con lo mismo sería genial. Gracias

about 4 years ago · Juan Pablo Isaza
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