Question: In a programming language like javascript, even though I don't make any a "returned" result in the end, but it still returns a value, "undefined".

In case of Sass, I know that @mixin does not "return" anything like @function. 

However, it "actually" returns nothing at all?

If it returns something actually (something like "undefined" or "void"), then what it returns?

Answer: 
I'm not sure I get the question but...

Mixins lets you create reusable CSS declarations (CSS content) 'called' using @include.

Unlike functions you can't assign mixins to variables as they do not produce a return value (throws an error).

```
@function fs($value){
   @return $value;
} 

@mixin fs($value){
   font-size: $value;
   ...
} 

.class {                 //      returns value  
   font-size: fs(16px);  // =>   16px;
}

.class {                 //      generates properties and values  
   @include fs(16px);    // =>   font-size: 16px; ...
}

$var: fs(16px);          //  will work (function)
$var: @include fs(16px); //  won't work (mixin)

```
Try to read this [article](http://thesassway.com/advanced/pure-sass-functions) and this [one](https://www.competa.com/blog/sass-mixins-and-functions/), you have some answer :

> 
> "They obtain full CSS rules where properties that are dynamic will be passed into arguments"
> 
> 
> 

So when you @include a @mixin in your code, it will return **CSS rules**.