Local var not overriding parameter with same name?

Question

It just blew my mind:

function f(x) {
    var x;
    console.log(x); // prints 100!
}

f(100);

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I always thought that whenever you declare a local variable, everything that was declared earlier is hidden behind it. Why var xdoesn’t it x undefined, as it was in other situations?

Compare with this code, which behaves the way I used:

var x = 100;
function f() {
    var x;
    console.log(x); // prints undefined as expected
}

f();

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Answer 1

For function code, parameters are also added as bindings to this environment record.

A specific parameter is xmatched to the current function. Since it is xalready declared in the function as a parameter var x,; the ad will be ignored.

More details: http://es5.imtqy.com/#x10.5

Answer 2

var, x . , undefined 100, .

Answer 3

MDN # var:

... JavaScript, .

, x f ( ), ( var) x redeclaration ( , f, ).

In the second example, it is xdeclared inside a function f. Therefore, when viewing the values xfor the transition to, logwe use this xinside f(and not external), and since we obviously did not give it a value (initializing it), its value will be undefined.