Reputation: 965
I am returning to javaScript after a long absence and I've just learnt about ES6 new features. And here it comes the first doubt.
Although I understand that a let
variable cannot be declared twice in the same block scope I came across this situation:
function tell(story){
let story = {
story
};
}
And it gives me a error:
Uncaught SyntaxError: Identifier 'story' has already been declared
Using var
this works just fine.
1) Does this mean that I will have to declare a variable with a different name of the function's parameter using let
(or even const
)? This can be very annoying in terms of variable naming.
In order to keep the same name I did:
function tell(story){
story = {
story
};
}
2) In this case the variable story
will belong exclusively to the scope of the function tell
?
Upvotes: 4
Views: 1349
Reputation: 21
"the var is completely ignored"
Not really, consider the following two examples:
1:
function someFn(a, b = () => a) {
console.log(a, b());
a = 2;
console.log(a, b());
}
someFn(3);
2:
function someFn(a, b = () => a) {
console.log(a, b());
var a = 2;
console.log(a, b());
}
someFn(3);
In the first example the output will be:
3 3
2 2
And in the second one:
3 3
2 3
So var
does matter.
Upvotes: 2
Reputation: 1074108
Using
var
this works just fine.
Not really, the var
is completely ignored, and you'd effectively have:
function tell(story){
story = {
story
};
}
If that's what you want, just leave the var
(and let
) off entirely (as above).
The error from let
addresses this pitfall, where you think you've created a (new) variable but haven't. You can't use let
(or const
) to declare a variable that's already declared in the same scope (either as a variable, or as a parameter, which is almost the same thing as a variable).
1) Does this mean that I will have to declare a variable with a different name of the function's parameter using
let
(or evenconst
)? This can be very annoying in terms of variable naming.
That's up to you. To keep doing what you were doing, just leave off var
, let
, and const
entirely, because again, the var
had no effect at all and you were assigning to the parameter, not a new variable. Or keep doing what you were doing (with var
), but understand that it's misleading. Or use a different name with let
or const
— there are many who believe changing the value of a parameter isn't best practice (personally I'm not one of them, but...) and they'd recommend a different variable instead.
2) In this case the variable
story
will belong exclusively to the scope of the functiontell
?
Yes, you're assigning to the parameter story
, which is local to tell
.
Upvotes: 6