How do constructor Function () arguments pass quoted parameters and unquoted parameters operate on local variables and external variables?

(function(){
        var call = "23";
       Function(console.log(call))(); //23
    })();

print is 23 

(function(){
        var call = "23";
       Function("console.log(call)")(); //
    })();

I don"t understand why the result of an unquoted parameter is different from that of an unquoted parameter.

Javascript
Apr.07,2021

Let's put it this way, first of all, make one thing clear. 

(function(){ 
    //code
})();

this is used to create closures, regardless of this.
then look directly inside.

Function is the prototype of all javascript functions, and you can create a function through the
Function (args) method.

in the first example, what is passed to the Function function is console.log (call)
in the closure, call = "23", console.log is executed immediately, so the result is actually
Function (undefined) (),

. For the second example, take a look at MDN, MDN that makes it clear that variables within the
clipboard.png
closure will not be referenced, so although a function

has been successfully created here 
 anonymous() {
console.log(call)
}

but it does not have the scope of call and will report a call is undefined error when it is executed immediately.

this question is mainly about this feature of Function

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