如何通过定义为函数的变量将对象名称作为参数传递给另一个函数？

Question

In my expanded code, I'm getting a TypeError and I think it's related to me not successfully passing an object as a parameter from one function, then through a variable defined as a function, and then on to the final function.

More specifically, I am calling the variable executeOnce (which is defined as a function) and passing the parameter options to it. This self-executing function then must pass along this options parameter to funB() (ie, funB(options) ) but I don't think the passed argument is making it to funB() . Hence the error.

What am I missing?

In the code below, it works if I change funB(options); to a string (ie, funB("options"); ), but I can't do this because in my expanded code I am passing various arguments in.

 const obj = { options: { spam: 4 }, }; function funB(options) { obj[options].spam = 6; // the console log below should print "6" but doesn't } var executeOnce = (function (options) { var executed = false; return function () { if (!executed) { executed = true; funB(options); // the function works if i change this to 'funB('options');' } }; })(); funA('options'); function funA(options) { executeOnce(options); } console.log('= ' + obj.options.spam) 

Answer 1

You are wrong at executeOnce , you use immediate function syntax to build a value for executeOnce variable (sound good), but executeOnce is a function without any parmas, you can see your return statement.

var executeOnce = (function (options) { // `options` - does not make sense
    var executed = false;
    return function () {
        if (!executed) {
            executed = true;
            funB(options);
        }
    };
})();

Try my way, return a function what has options is a parameter.

 var executeOnce = (function () {
    var executed = false;
    return function (options) {
        if (!executed) {
            executed = true;
            funB(options);
        }
    };
})();

Answer 2

  const obj = { options: { spam: 4 }, }; function funB(options) { if(obj[options]==undefined) obj[options]={}; obj[options].spam = 6; } var executeOnce = (function (options) { var executed = false; return function (options) { if (!executed) { executed = true; funB(options); // the function works if i change this to 'funB('options');' } }; })(); funA('optionsA'); function funA(options) { executeOnce(options); } console.log(obj) 

Answer 3

I think you might be creating one more function than you need, but that doesn't break anything. The issue seems to be that the returned function (closure) doesn't know about options , which can be solved by passing options in as an argument when the returned function is called.

I gave the returned function a name, newFunc , and also added print statements throughout the script to clarify what's happening at each step (which made debugging way easier.)

 // Main const obj = { options: { spam: 4 } }; funA(obj.options); console.log("finally..."); console.log('obj.options.spam = ' + obj.options.spam); // Functions function funB(options){ console.log("funB got..."); console.log(options); options.spam = 6; console.log("funB set options to..."); console.log(options); } function executeOnce(options){ // context for closure (IIFE) console.log("executeOnce got..."); console.log(options); var executed = false; var newFunc = function(options){ console.log("newFunc got..."); console.log(options); if(!executed){ executed = true; funB(options); } }; return newFunc(options); }; function funA(options){ console.log("funA got..."); console.log(options); executeOnce(options); }