Why the f-I does not return a value?

Question

Why the f-I does not return a value?

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const test = (arg) => { return arg }; const final = (arg) => test => { test(arg) } final("test");

Code
As I understand it, the f-th final we pass the argument arg and pre-generated f-th test, and in the body of f-and final we call f-u test(). It?
For some reason, I do not receive values from the call to final("test");

asked Apr 12, 2019 by Nikulio | 32 views

3 Answers

delphinpro · Answer 1 · 2019-04-16T06:27:22+0000

rewrite the old way, and you will see that there is generally some crap...

const test = function(arg) { return arg; } const final = function(arg) { return function(test) { test(arg); } } final('test');

Actually should return a function, the argument is lost (not used) well, etc.

UPD
Launched your code in the console. Actually a function and returns

Jumandjilos · Answer 2 · 2019-04-16T06:27:22+0000

Remove the excess.

const test = (arg) => { return arg }; const final = (arg) => test(arg) final("test"); /* test */

sergiks · Answer 3 · 2019-04-16T06:27:22+0000

You have no return

const final = (arg) => test => { return test(arg) }

Or you can do so:
const final = (arg) => test => test(arg);

Why the f-I does not return a value?

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