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JavaScript JavaScript Loops Working with 'for' Loops The Refactor Challenge – Duplicate Code

italiandadx2
italiandadx2
3,566 Points
Posted by italiandadx2
italiandadx2
3,566 Points

Problem with the value parameter

Wouldn't the value past into the function be the same for each rgb number...? Meaning when you call the function and pass it a value of say 10... wouldn't the color returned be rgb(10, 10, 10)...? I guess you still get a random color for each circle but each color would have the same value for for r, g and b... rgb(10, 10, 10), rgb(127, 127, 127), etc...

italiandadx2
italiandadx2
3,566 Points
italiandadx2
italiandadx2
3,566 Points

Thank you! Took me a minute to get what you were saying... I didn't realize that when you pass a function as an argument that the function would be called independently each time it's referenced... Thank you again!!!

3 Answers

Peter Vann
Peter Vann
36,427 Points

Hi!

If you pay close attention to the video right around 5:00, you will notice that the value is actually the randomValue function being passed in.

The give-away being this line of code:

const color = `rgb( ${value()}, ${value()}, ${value()} )`;

Which tells you that value() has to be an executable function (randomValue, specifically, which will randomly generate a fresh, distinct number each time it is called - in this case, three times right in a row).

Does that make sense?

I hope that helps.

Stay safe and happy coding!

Yazeed Hani
Yazeed Hani
5,002 Points
Yazeed Hani
Yazeed Hani
5,002 Points

So you're not actually passing the randomValue function into the randomRGB function in the for loop. You are just passing in the word "randomValue" as an argument to the randomRGB function. The parameter "value" in the function defintion of randomRGB will store the word "randomValue" and pass it into the ${value()} placeholders.

So, technically, ${value()} is literally ${randomValue()}.

daniel barreto
daniel barreto
4,145 Points
daniel barreto
daniel barreto
4,145 Points

When you pass " randomValue" to "randomRGB" you are only passing the variable "randomValue" which

happens to be a function. " randomValue" exists inside "randomRGB" as a term called "value".

when ${ value() } is ran it is invoking the "randomValue()" function. the parenthesis make the function perform

its operation.