function aFun(){????function _(arg){????????this.arg = arg;????}????_.prototype.testFun = function(){????????console.log('this is prototype testFun')????};????_.testFun2 = function(){????????console.log('this is testFunc2');????}????return _;};var afun = new aFun('123');console.log(afun instanceof aFun)afun.testFun2(); //可以調用afun.testFun(); //不是在原型鏈上么?,為啥不可以調用???,var aFun2 = (function (){????function _(arg){????????this.arg = arg;????}????_.prototype.testFun = function(){????????console.log('this is prototype testFun')????};????_.testFun2 = function(){????????console.log('this is testFunc2');????}????return _;})();var afun2 = new aFun2('ddd');console.log(afun2 instanceof aFun2)afun2.testFun2(); //不可以調用,為啥?afun2.testFun(); //可以調用,為啥可以?(我要瘋了)
javascript 函數返回值的問題,請高手解答
fshang
2018-08-28 13:29:49