Fourth Way, using Function.prototype.bind() #16

jatinarr · 2021-01-23T07:59:18Z

Create the function to be called outside the loop. While sending it as a callback, bind it to null with an initial argument as i.

Why does it work? Basically, we are not sending the original function as setTimeout's callback but actually sending the return value from the bind operation, and that return value is the copy of the function. Also, bind allows us to not immediately call the function like .call() or .apply() does.

Code:

function display(x){
  console.log(x)
}
for (var i = 0; i < 4; i++) {
  setTimeout(display.bind(null,i), 0)
}

Create the function to be called outside the loop. While sending it as a callback, bind it to `null` with an initial argument as `i`. Why does it work? Basically, we are not sending the original function as `setTimeout`'s callback but actually sending the return value from the bind operation, and that return value is the copy of the function. Also, bind allows us to not immediately call the function like .call() or .apply() does. Code: ```js function display(x){ console.log(x) } for (var i = 0; i < 4; i++) { setTimeout(display.bind(null,i), 0) } ```

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Fourth Way, using Function.prototype.bind() #16

Fourth Way, using Function.prototype.bind() #16

jatinarr commented Jan 23, 2021

Fourth Way, using Function.prototype.bind() #16

Are you sure you want to change the base?

Fourth Way, using Function.prototype.bind() #16

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jatinarr commented Jan 23, 2021