在JavaScript中检查没有hasOwnProperty的哈希键的存在

Question

Given this example data:

let foo = [1,2,2,3,5,3]
let seen = {}

What is the difference between the following ways to check the existence of a key? Can I run into some hiccups using the first way?

foo.filter(function(item) {
    return seen[item] ? false : (seen[item] = true);
});

vs.

foo.filter(function(item) {
    return seen.hasOwnProperty(item) ? false : (seen[item] = true);
});

Answer 1

First off, as shown in the question, both are incorrect because you're not using the return value of filter (a new, filtered array). Use forEach or (on modern systems) for-of in that case. But perhaps you actually are using it and just didn't show that in your question.

Addressing your main question:

The first way will incorrectly assume something wasn't seen if it's falsy . The falsy values are 0 , "" , NaN , null , undefined , and of course false . (All other values are truthy .) Your second way will handle those correctly.

Another difference between your two checks is that the first gets the property regardless of where it is in the prototype chain; the second only looks at the object itself (which is probably what you want). For your example object, this really only matters for properties provided by Object.prototype such as valueOf , but...

You might also look into Set or Map .

Answer 2

Short answer : there are no problems in this case for not using Object.hasOwnProperty() [wiki]

When would it be a problem?

 let seen = {} console.log(seen["constructor"]) // function Object() { [native code] } 

as you can see this will also hold other strings for inherited properties like constructor of an Object. Since you are only dealing with numbers there a no such properties on Object.

Answer 3

In addition to what TJ said.

hasOwnProperty will not search the prototype of the object while the bracket or dot notation will. See this example

 Object.prototype.foo = 1; var baz = {"bar": 1} if (baz.hasOwnProperty('foo')) { console.log('has foo'); return true } if (baz.bar) { console.log('has bar'); // has bar return true; }