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Math.min.apply(0, array) - why?

开发者 https://www.devze.com 2022-12-31 21:32 出处:网络
I was just digging through some JavaScript code (Raphaël.js) and came acro开发者_如何转开发ss the following line (translated slightly):

I was just digging through some JavaScript code (Raphaël.js) and came acro开发者_如何转开发ss the following line (translated slightly):

Math.min.apply(0, x)

where x is an array. Why on earth would you do this? The behavior seems to be "take the min from the array x."


I realized the answer as I was posting my own question: This is the most succinct way of taking the min of an array x in JavaScript. The first argument is totally arbitrary; I find the 0 confusing because the code intuitively means "Take the min of 0 and x," which is absolutely not the case. Using the Math object makes more sense for human-readability, but the Raphael.js authors are obsessed with minification and 0 is three bytes shorter.

See http://ejohn.org/blog/fast-javascript-maxmin/

For readability's sake, I'd strongly urge people to stop doing this and instead define a function along the lines of

function arrayMin(arr) { return Math.min.apply(Math, arr); };


The reason is this:

  • Your input x is an array

  • The signature of Math.min() doesn't take arrays, only comma separated arguments

  • If you were using Function.prototype.call() it would have almost the same signature, except the first argument is the this context, i.e. who's "calling"

    • Example: Math.min.call(context, num1, num2, num3)
  • The context only matters when you refer to this inside the function, e.g. most methods you can call on an array: Array.prototype.<method> would refer to this (the array to the left of the 'dot') inside the method.

  • Function.prototype.apply() is very similar to .call(), only that instead of taking comma-separated arguments, it now takes an array after the context.

    • Function.prototype.call(context, arg1, arg2, arg3)
    • Function.prototype.apply(context, [arg1, arg2, arg3])
  • The 0 or null that you put in as the first argument is just a place shifter.


The proper and less confusing way to call this is: Math.min.apply(null, array)

If parameter is unused setting it to null makes code more readable than setting it to 0 or Math

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/apply


Here is an example where we can get the MAX and MIN timezone offsets for the given year for a timezone. It returns the two changes for the year to account for DST and ST. (i.e. ST = 300 an DST = 360 for the timezone in question).

   var arr = [];
   for (var i = 0; i < 365; i++) {
       var d = new Date();
       d.setDate(i);
       var newoffset = d.getTimezoneOffset();
       arr.push(newoffset);
   }
   var DST = Math.min.apply(null, arr);
   var nonDST = Math.max.apply(null, arr);
   tw.local.stInt = DST
   tw.local.edInt = nonDST
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