创建ES6-class实例后，是否可以立即将实例绑定到非静态方法? [英] Can I bind the instance to non-static method immediately after creating instance of ES6-class?

问题描述

以某种方式将前两行组装成一个吗?我不愿意强迫MarkupPreprocessingHelper的用户写两行...

Is it some way to assemble first two rows into one? I am not feeling comfortable to force users of MarkupPreprocessingHelper write two rows...

let markupPreprocessingHelper = new MarkupPreprocessingHelper(config);
let preprocessTemplates = markupPreprocessingHelper.takeCareAboutMarkupPreprocessing.bind(markupPreprocessingHelper);

gulp.task('Development run', gulp.series(
   preprocessTemplates,
   // ...
));

推荐答案

如果制作该函数的反弹副本并将其另存为实例属性，则可以将其传递出去，并且用户无需手动绑定它:

If you make a rebound copy of the function and save it as an instance property, you can then pass it around and users won't need to bind it manually:

function someClass(name){
    this.name = name
    // make a prebound copy of myFunction
    this.preBound = this.myFunction.bind(this)
}

someClass.prototype.myFunction = function(){
    console.log(this.name)
}

let p = new someClass("Mark")

// now you can pass a reference of it around without losing the binding
let fn = p.preBound
setTimeout(fn, 500)

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