variable.constructor Javascript.

[mrchow19910319]

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()

??

 

 

 

[IAmAndre]

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.

[Franck]

This look like the good old X,Y problem

[mrchow19910319]

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())

[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.