General Curry Function

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General Curry Function

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#1 by (1 votes)

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Could someone explain how this function works? I'm studying currying (I do not know what Spanish translation you will have) and in a part of the book it teaches this function as a general curry function (or a general curry function).

I find it hard to understand the methods that it calls over the array in the second and fourth lines. The call method is not unique to functions? Should not pass a start and end arguments of slice ? Also, it is not assumed that the arguments object can not be called from nested functions?

function curry(func) {
  var fixedArgs = [].slice.call(arguments,1);
  return function() {
    args = fixedArgs.concat([].slice.call(arguments))
    return func.apply(null, args);
  };
}

To understand it, the author gives an example of its use. Still, I can not fully understand how it works. But I transcribe it here:

This can now be applied to a more standard divider () function that returns the result of dividing its two arguments:

function divider(x,y) {
  return x/y;
}

divider(10,5);
<< 2

The curry () function can be used to create a more specific function that finds the reciprocal of numbers:

reciprocal = curry(divider,1);
<< function () {
      args = fixedArgs.concat([].slice.call(arguments))
      return func.apply(null, args);
    }

reciprocal(2);
<< 0.5

In case someone helps you to understand it and be able to answer my question, I leave the book and its author below: JavaScript: Novice to Ninja - Darren Jones

EDIT : I begin to understand it. In line 2, call ( call ) the function slice and pass it as arguments the object arguments (as it is the first argument and as the function indicates call will be the object on which the function is applied, that is, where the internal structure of slice has a this. ) and 1, that being the second index, the function that is passed as an argument to curry will be skipped. Then all the rest of the arguments passed to curry will be added to an array that is the variable fixedArgs . When the function curry was called, it will return another function that are the 4 and 5 line. In this function, we call again slice through call and since there is only one argument (which is the object arguments ), all the arguments passed to this new function will be added to the variable fixedArgs ( by concat and then to variable args . Finally the function will return the call (through apply ) of the function originally passed as argument of curry with all arguments within args .

And to generalize and define it in a more theoretical way: The curry function, as expressed here, allows to call a function without the totality of its arguments, which will then return another function that waits for the remaining arguments in its invocation.

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asked by Federico Grandon Soporsky 24.05.2017 в 03:59

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1 answer

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score 1 · Answer 1

example

I will try to explain it with an example. Suppose that you already had programmed a function that abbreviates a text to n characters (to print something reduced), receives the number n and the text t. It could be something like this:

function abreviar(n, t){
    if(t.lenght<=n) return t;
    t = t.replace(/\s+/g, ''); // quita espacios
    if(t.lenght<=n) return t;
    return t.substr(0,n-3)+'...';
}

Suppose now that we want to reduce an array of names to 10 characters with the abreviar function. It can be done without currifying:

var a = ['John Nash', 'Jhon Von Neumman'];
var aa = a.map(function(nombre){
    return abreviar(10, nombre);
});

Currifying is made that easy:

var a = ['John Nash', 'Jhon Von Neumman'];
var aa = a.map(curry(abreviar, 10));

In other words, apply the function abbreviate with its first parameter in 10 [ie currificar abbreviate] to each element of the array [ie map].

answers to specific questions

1) yes call is almost exclusive to the functions. But slice is a function. (I say almost because you can add to any object a function called call).

2) No. slice can receive beginning and length or only principle (and understands that it is until the end. call receives the object to apply the function (this) and the parameters it receives. , b, c) equals to ofcall (or, a, b, c)

3) arguments can be consulted from internal functions, what happens is that arguments refers to the most internal function.