ReactJS | Why super(props) and bind() in constructor

    This article tries to answer some WHY questions making me confused at the first time that I learn ReactJS about super(props) and bind method in constructor.

    This post just explain some lines of code based on JavaScript knowledge. I do not go to deep theories, therefore, you need to read it through recommended articles below in case you are not familiar with these definitions before.

Requirements knowledge:

1. Why should we bind methods in constructor?

    image.png

    As we have already known:

    this in a normal function is determined by how it is called, no matter where it is written.

    In the code above, handleRemoveAll() is fired by the click event of <button>. Therefore in this case, this is not point to class Options but to event.target.

    For that reason, using bind method to force this on handleRemoveAll method points to the class Options.

2. Why do we have to call super() on constructor?

    Some questions that spring to my mind when I see super() in constructor at the first time: 'Why do we call super() on constructor although everything seems okay without constructor before ?'

    I start to search and common answers on Internet say that:

    We cannot use this in constructor until super() is called

    But WHY?

    So let jump into this example with me.

    Imaging that we can use this in constructor without calling super()

    image.png => The code above is ok, everything works well.

    But what happens when we access this.name in actionType method

    image.png

    Well, definitely it will come into error. Because Dog class does not have name property. We only can use this.name when call super first to inherit name property from parents class.

    Therefore,

    JavaScript enforces that if you want to use this in a constructor, you have to call super first

    This makes sure the program run smoothly at all times.

3. Why 'props' is passed to super()?

    Once again, the answer on the Internet is:

    Passing props to super helps us use this.props in constructor, otherwise, it will become undefined

    WHY? Why we can call this.props outer constructor but it turns into undefined inside the constructor?

    On ReactJs Document:

    The constructor for a React component is called before it is mounted (Inserted into the tree)

    Therefore, this ensures this.props is set even before the constructor exits.

    That's all I want to share with you in this post 😀

    Reference:

  1. Why Do We Write super(props)
  2. Why do we have to ‘bind’ our functions in React apps?

Nguồn: Viblo

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