If re-declaration with var will have effect on exi

问题:

const show1 = function(x, y = () => {x = 2; return x;}) {
    let x = 3;
    console.log(y());
    console.log(x);
};
show1();

const show2 = function(x, y = () => {x = 2; return x;}) {
    x = 3;
    console.log(y());
    console.log(x);
};
show2();

const show3 = function(x, y = () => {x = 2; return x;}) {
    var x = 3;
    console.log(y());
    console.log(x);
};
show3();

output

show1: Uncaught SyntaxError: Identifier 'x' has already been decalred;
show2: 2 2
show3: 2 3

Question

I was informed that there is a temporary dead zone where parameter variables are declared and initialized. See https://exploringjs.com/es6/ch_variables.html#sec_parameters-as-variables. So there are two scopes here, one is the parameter scope and the other is function scope.

1. From the error in show1, I thought that there is a x variable already declared in this function.
2. According to Redeclaring a javascript variable. The re-declaration won't do anything to x(with var). Why the results of show2 and show3 are different.

I posted the same question here If the variables are already declared at the beginning of function execution which was masked as duplicate. But I couldn't find anything useful for my doubt.

回答1:

show1 throws an error because variables declared with let or const cannot have any other variables with the same name initialized in that block (whether in an argument list or with const / let / var).

---

Variables referenced by default parameters have odd scoping rules. Every argument essentially creates another block that that argument name can be defined in. So

const show3 = function(x, y = () => { x = 2; return x; }) {

is somewhat like (forgive the psuedo-code):

const show3 = < function >{
  let x = firstArg;
  {
    let y = secondArg === undefined ? () => { x = 2; return x; } : secondArg;
    {
      // function body

When you simply assign to a variable name in the argument list, that argument will be overwritten. But when you use syntax to initialize a new variable (with var), you've created another binding for that variable name, which is only visible inside the function body:

const show3 = {
  let x = firstArg;
  {
    let y = secondArg === undefined ? () => { x = 2; return x; }
    // the x referenced in the above line references the outer x, the first argument
    {
      // function body

      // since "var x" is declared here, any references to "x" in this block
      // will only reference the inner x
      // but not the outer x
      var x = /* something */

So, your show2 is reassigning the parameter named x to 3 in the first line of the function body. In contrast, show3 is creating a new variable binding with the same name of x, while the y function's reference to x is referencing the argument x, which is different.

回答2:

Why the results of show2 and show3 are different.

let's evaluate your code by this way

const show2 = function(x, y = () => {x.value = 2; return x;}) {
    x = {name: "from argument", value: 3};
    console.log(y());//{name: "from argument", value: 2}
    console.log(x);//{name: "from argument", value: 2}
};
show2();

const show3 = function(x, y = () => {if(!x){x = {name:"from function", value: -1}}x.value = 2; return x;}) {
    var x = {name: "from var", value: 3};
    console.log(y());//{name: "from function", value: 2}
    console.log(x);//{name: "from var", value: 3}
};
show3();

const show4 = function(x, y = () => {if(!x){x = {name:"from function", value: -1}}x.value = 2; return x;}) {
    var x = {name: "from var", value: 3};
    console.log(y());//{name: "from outside", value: 2}
    console.log(x);//{name: "from var", value: 3}
};
show4({name:"from outside", value: -1})