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Make call to super() implicit within constructor #5910
Description
Calling super() within a constructor is a bit redundant and annoying... can we have it called implicitly if call to super is omitted?
For example this
class Manager extends Employee {
constructor() {
this.title = "The Manager";
}
}should be equivalent to
class Manager extends Employee {
constructor() {
super(); // call to default constructor added implicitly
this.title = "The Manager";
}
}(C# also does a similar thing and it's a nice feature).
Activity
coreh commentedon Dec 3, 2015
What would be the default behavior for the parameters? Would they be passed on to the super class constructor if the signature was the same?
nippur72 commentedon Dec 3, 2015
Marco Buono (@coreh) that would be a possibility, since you can always override it by making an explicit call to
super(). It would do no harm to existing code.GoToLoop commentedon Dec 3, 2015
Gonna add that like C#, Java does the same too for empty super constructors! :D
mhegazy commentedon Dec 3, 2015
TypeScript is and should be a super set of JavaScript ES6 semantics has no implicit call for constructors, if TS supports that, a TS/ES6 class are not identical in behavior.
GoToLoop commentedon Dec 3, 2015
As long as some handy feature isn't detrimental to ES6 compatibilty, I don't see why it should be thrown away.
This proposal merely asks for TS' compiler to match some of the smartness of C# & Java compilers.
It won't affect codes which don't use this saving grace feature.
Since TS is a compiler/transpiler it can do tricks that raw JS can't after all.
Many C# & Java programmers would be attracted to JS world if TS would mirror many of C#/Java features where it's feasible to do so.
nippur72 commentedon Dec 3, 2015
peraphs there could be a compiler flag like #274 to turn on and off features like this one.
nippur72 commentedon Dec 4, 2015
@jasonwilliams200OK the new behaviour would be:
super()call and the class has a default constructor, instead of throwing an error, make a call tosuper()as the first thing in the constructor.Your special case would be covered by calling super explicitly. (And btw, is it possible to refer to
thisbefore super?)RyanCavanaugh commentedon Dec 5, 2015
Note that this is disallowed in ES6. You cannot access
thisuntil after thesupercall.mhegazy commentedon Feb 19, 2016
As mentioned above, we would like to keep the semantics matching JS, and issue warnings to help guide the user the correct behavior.