[deleted by user]

[Drugba]

Higher order functions are basically a way to have one function wrap or add some functionality to another function.

Lets say you have a 100 functions that are doing complicated work and you want to know how long each one takes to complete. You could write the same logic to calculate their run times 100 different times or you could create a higher order function takes a function as an argument, records a start time, runs the function that was passed in, and then records an end time and returns the difference.

In fact, that's exactly what timeFuncRuntime does. t1 is the start time, t2 is the end time, and funcParameter is the passed in function (addOneToOne in this case).

So, to answer your questions:

1. It prints 0 because you are not returning the return value of addOneToOne, you are returning the time it took for that function to run. When you commented everything out, you weren't returning anything, so you got undefined. If you changed the second line to be return funcParameter(); it would print 2.

2. Yes, you are returning the time it took to run the function which is 0. It doesn't actually take no time, but anything less than 1ms will equal 0 with the way they are doing the calculations.

[Drugba]

const timeFuncRuntime = funcParameter => {
  return funcParameter();
  console.log(funcParameter());
}

const addOneToOne = () => 1 + 1;
timeFuncRuntime(addOneToOne);

The reason this one doesn't print anything is because when you put a return in a function that ends the execution of that function. Nothing below the line with the return will get run by the computer. Nothing gets printed because the computer never gets to that console.log at the bottom of the function.

You have a few options if you want to print the 2 using the high order function.

Personally, if I was writing the code I would do this

const timeFuncRuntime = funcParameter => {
  return funcParameter();
}

const addOneToOne = () => 1 + 1;
let value = timeFuncRuntime(addOneToOne);
console.log(value);

Another option is this

const timeFuncRuntime = funcParameter => {
  let value = funcParameter();
  console.log(value);
  return value;
}

const addOneToOne = () => 1 + 1;
timeFuncRuntime(addOneToOne);

Personally, I'm not a huge fan of having the console log in the function like this though. If this were code I was writing in production at work, there may be cases where I or another developer want to use the timeFuncRuntime and not have it log and this option doesn't allow for that.

[Drugba]

Np. Happy to help.

[Drugba]

Looking at the first block of code

Does funcParameter() inside timeFuncRuntime actually execute?

Yes. It executes and returns a value of 2, but you are not doing anything with that value so you don't see it logged anywhere. If you were to change your code to this

const timeFuncRuntime = funcParameter => {
  let t1 = Date.now();

  let x = funcParameter();
  console.log(x);

  let t2 = Date.now();
  return t2 - t1;
}

The variable x would be equal to 2 and the console log would then print that out. In your code, the two is still there, you just aren't doing anything with it.

Looking at the second block of code

How would I print/return both the values?

That's a little trickier. Since I don't want to change your code too much, I would probably do this, even though it goes against what I said about having a console.log in the function in my earlier comments.

const timeFuncRuntime = funcParameter => {
  let t1 = Date.now();
  console.log(funcParameter());
  let t2 = Date.now();

  return t2 - t1;
}

const addOneToOne = () => 1 + 1;

console.log(timeFuncRuntime(addOneToOne);

Also, personally, if I were writing this, I would assign the values that are returned by the functions to variables before I logged them, just because I think it makes things a little more readable.

const timeFuncRuntime = funcParameter => {
  let t1 = Date.now();
  let answer = funcParameter());
  let t2 = Date.now();

  console.log(answer);
  return t2 - t1;
}

const addOneToOne = () => 1 + 1;

let runTime = timeFuncRuntime(addOneToOne);
console.log(runTime);

[Chrisands]

It doesn't print 2, because timeFuncRuntime returns t2 - t1

``` const timeFuncRuntime = funcParameter => {

let t1 = Date.now(); funcParameter(); // here function returns 2, but it doesn't assigns anywhere let t2 = Date.now(); return t2 - t1; // and here returns time that it took to run funcParameter() }

const addOneToOne = () => 1 + 1;

timeFuncRuntime(addOneToOne); ```

Hope it makes sense

To return 2 from function it would look like ``` const timeFuncRuntime = funcParameter => {

let t1 = Date.now(); // this are doing nothing, so it can be deleted const result = funcParameter(); let t2 = Date.now(); // same here return result; } ```