CODEWITHSUNDEEP
javascriptconstructorcall
Huibin Zhang
Huibin Zhang

Reputation: 1110

Why the code is put after the return statement, will it get executed?

I don't understand why the Function.prototype.call() can be used this way? As far as I know if a function returns the code after that will not get executed. Did I missed something here? 

function Product(name, price) {
  this.name = name;
  this.price = price;

  if (price < 0)
    throw RangeError('Cannot create product 
                 "' + name + '" with a negative price');
  return this;
}

function Food(name, price) {
  Product.call(this, name, price); // if the function returns here why put this.category after this statement?
  this.category = 'food'; // will this ever get executed? 
}
Food.prototype = Object.create(Product.prototype);

var cheese = new Food('feta', 5);

I understand that both the Product, Food are constructors and we can use call to chain constructors for an object, similar to Java. But why don't put the statement 

this.category = 'food';

before 

Product.call(this, name, price);

Upvotes: 0

Views: 85

Answers (1)

Bergi
Bergi

Reputation: 664970

As far as I know if a function returns the code after that will not get executed.

Yes.

Did I missed something here?

return works local, and ends only the current function call. The Food function does not return before setting the .category property.

Btw, the return in Product is unnecessary as constructors don't need to explicitly return.

Upvotes: 1

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