ES6带有循环的箭头符号

Question

Consider the following bit of code:

f=(m,c)=>{m?c()&f(--m,c):0}

(thanks to zzzzBov for this little nugget)

which is a "for"-less loop

and the following:

a=b=>b+1

Given these two snippets, and the fact that:

z = 0; f(10,a(z));

which I would expect would result in z equating to 10, but instead returns in the JavaScript console the following "TypeError: c is not a function" , how would one go about altering this code to ensure that the loop goes ahead, without having to resort to a while or for loop?

I'm asking this as a matter of education purposes... Hopefully I can get some insight into what can be done...

Answer 1

The function f is taking 2 arguments: m , the number to iterate, and c , the function to be called m times. This means that the second argument, c should be a function. For example:

f=(m,c)=>{m?c()&f(--m,c):0}


f(15, function() {
 console.log("Hello")
})

This will iterate through the c function 15 times, calling console.log 15 times.

Of course, to achieve what you wanted in the second bit, you could use this:

z=0, f(10,()=>z++)

This would be a regular arrow function to increase z by 1

Take a look at the code on babel

Hope I could help!

Answer 2

It sounds you are looking for a folding function (like array reduce ), not a simple "looping" function that only executes side effects. With that current function, which desugars f(5, c) to effectively c(); c(); c(); c(); c(); c(); c(); c(); c(); c(); you would need to do

let z = 0;
f(10,()=>{ z = a(z) });

If however you want to make a function that repeatedly applies a function, like a(a(a(a(a(…))))) , you would need to write

let times = (n, f, s) => n>0 ? times(n-1, f, f(s)) : s;

so that you can do

let a = b=>b+2
times(5, a, 0) // 10