I was trying to learn about closures and this the following code:
var one = 1, two = 2, three = 3; function bom(one, two, three) { console.log(one, two, three); function b(one) { console.log(`From closure ${one}`); }; b(); }; bom(1, 2, 3);
However, the inner function has no access to the outer function variables.
Can anyone explain to me why?
Thanks.
What @Pointy said is correct, you have done what's called variable shadowing on the one
variable, meaning you have essentially overwritten the reference to the outer one
variable inside the b
function scope. It will reference the local one
variable defined in its parameter list instead.
That being said, this really isn't an example of a closure, though, as the function definition and calling context are within the same scope. It would be a better example of closure if you returned the function b
and then executed it later. That proves the bom
scope lives on within the closure of the b
function scope.
Here's an example:
var one = 1, two = 2, three = 3; function bom (one,two, three){ console.log(one,two,three); return function b(){ console.log(`From closure ${one}`); }; }; let myClosureFn = bom(1, 2, 3 ); myClosureFn();
If b and bom got no parameter named one they could use the one that's defined as var.
var one = 1,
two = 2,
three = 3;
function bom (two, three) {
console.log(one, two,three);
function b () {
console.log(`From closure ${one}`);
};
b();
};
bom(2, 3 );
output is :
1 2 3
From closure 1
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