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variable.constructor Javascript.

mrchow19910319
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Go to solution Solved by 0x21,

Dog() is a call to the function without passing a parameter for the name.

Dog is the name of the function.

 

What you are doing in this case is:

console.log(new_variable.constructor, Dog(), new_variable.constructor === Dog());
// [Function: Dog] undefined false

You can read more about constructors here.

let’s say that there is a constructor function here:

function Dog(name){
	this.name = name;
}

let new_variable = new Dog("puppy");

console.log(new_variable.constructor);
console.log(new_variable.constructor===Dog);

the 1st console.log will return you a function , because .constructor will return you a reference of the constructor function, following this logic if .constructor will return you a function, shouldn’t be the second statement becomes:

 

new_variable.constructor===Dog()

??

 

 

542493407_ScreenShot2018-10-16at6_12_00PM.png.429aecd3bb04885aef4fbde7090e7167.png

 

If it is not broken, let's fix till it is. 

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I'm really not sure what you're trying to achieve, but what I can tell for sure is that "===" is a comparison operator and will only return a Boolean value.

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1 hour ago, IAmAndre said:

I'm really not sure what you're trying to achieve, but what I can tell for sure is that "===" is a comparison operator and will only return a Boolean value.

why it is if(newvariable.constructor===Dog) instead of if(newvariable.instructor===Dog())

If it is not broken, let's fix till it is. 

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Dog() is a call to the function without passing a parameter for the name.

Dog is the name of the function.

 

What you are doing in this case is:

console.log(new_variable.constructor, Dog(), new_variable.constructor === Dog());
// [Function: Dog] undefined false

You can read more about constructors here.

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