http://www.2ality.com/2015/02/es6-cl...structor_calls
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Цитата:
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In a derived class, you must call super() before you can use this.
Implicitly leaving a derived constructor without calling super() also causes an error.
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http://www.2ality.com/2015/02/es6-cl...tanc e_object
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Цитата:
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In ES6, it [an instance] is created in the base constructor, the last in a chain of constructor calls.
this originally being uninitialized in derived constructors means that an error is thrown if they access this in any way before they have called super().
If a constructor returns implicitly (without a return statement), the result is this. If this is uninitialized, a ReferenceError is thrown. This protects you against forgetting to call super().
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Да, так что я попутал - не вызывать super в ребенке тоже нельзя.
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Посредством. По средствам только живут.
Этому форуму жутко не хватает тега hr, да.