我已经声明了一个回调函数，但我想调用它……但是失败了……如何直接调用它？

Question

类的编码函数...以及类中的定义...该函数在开始时就被调用...如何防止这种情况并像在函数测试中那样调用它....给出了所描述的错误？

...

// ==========================================================================================
// Action class - coded for onMsg to bring it to work....please change if needed
// ==========================================================================================
let action = new class Action {
    constructor() {
        this.msgFunction = (msg) => { msg };
        this.msg = { message: 'Hello!', to_user: 'PeterPan', from_user: 'AlexanderTheGreat' };
        this.onMsg = (MsgFunction) => { MsgFunction(this.msg) };
    }
}

// ==========================================================================================
// action functions: directly called at start - please NOT change
// ==========================================================================================
//let totalTokens = 0;
action.onMsg(function(msg) {
    console.log('==== onMsg ====');
});

// ==========================================================================================
// This is what I want: call the onMsg functions at any time!
// got error:
//      index.js:8 Uncaught TypeError: MsgFunction is not a function
//          at Action.onMsg (index.js:8)
//          at test (index.js:22)
//          at <anonymous>:1:1
// when I write "test()" in the console!
// ==========================================================================================
function test() {
    console.log('==== test ====')
    action.onMsg(action.msgFunction(action.msg));
}

...

Answer 1

在JavaScript中，您可以将对函数的引用作为参数传递。在您的代码中，第8行（错误所在）表示您正在传递的参数（test()的第2行）

例如：var myFunc = action.msgFunction是对函数的引用，因此您以后可以调用它。而action.msgFunction(action.msg)是此函数返回的值。

您的代码应如下所示：

function test() {
    console.log('==== test ====')
    action.onMsg(action.msgFunction);
}