CODEWITHSUNDEEP
javascript
Sadid Khan
Sadid Khan

Reputation: 1985

difference between string object, and primitive string

my problem is described below with example number-1:

 var myString = new String('foo');

if i use console.log(myString); the output is String { 0="f", 1="o", 2="o"}

and number-2:

var myString = new String();
myString = "foo";

here console.log(mystring); prints just foo

here what is the difference between number-1 and number-2 ? why the ouput is different ?

Upvotes: 9

Views: 327

Answers (3)

T.J. Crowder
T.J. Crowder

Reputation: 1074218

This statement:

var myString = new String('foo');

...creates a string object initialized with the characters f, o, and o.

This statement:

var myString = new String();

...creates a string object with no characters, but that doesn't matter because this statement:

myString = "foo";

...throws that string object away and replaces that variable's value with a new primitive string with those characters. The net result is exactly the same as:

var myString = "foo";

The reason the output from console.log is different is that the browser providing the console.log is trying to make it clear that one of them is an object and the other is a primitive.

Somewhat confusingly, JavaScript has both string objects and string primitives. (It also has number objects and number primitives.) There is almost never any reason to use new String, which creates a string object; just use the primitive (in your case, a literal) instead. Conversely, there are very good reasons not to use string objects, such as this:

console.log(new String("foo") === new String("foo")); // "false"
console.log(new String("foo")  == new String("foo")); // "false"
console.log("foo" === "foo");                         // "true"

Because string objects are objects, == and === compare object references, not the sequence of characters. While there may be some edge cases where that's what you want, 99.9% of the time, what you really want is to compare the values. The good news is that:

console.log("foo" == new String("foo")); // "true"

...but that won't be true if you use ===, which doesn't do any type coercion.

You might ask:

So if var myString = "foo" returns a primitive, not an object, how is it that myString.toUpperCase() and such work?

Good question: The answer is that the primitive is automatically promoted to an object so that we can make the function call. (In theory; in practice, implementations are smarter than that.)

Upvotes: 14

bbill
bbill

Reputation: 2304

console.log(typeof s_prim); // Logs "string"
console.log(typeof s_obj);  // Logs "object"

The difference is subtle, especially because methods called on string primitives are wrapped by string object methods, so functionally they are close to the same.

As pointed out in the MDN reference, this may matter on eval() calls. Eval will return the object "2 + 2" if called on the string object, and 4 if called on the primitive.

Upvotes: 2

Liviu T.
Liviu T.

Reputation: 23664

The best way to understand is by running these 2 statements

console.log(typeof 'foo'); // -> "string"

console.log(typeof new String('foo')); // -> "object"

Also the second statement you are reassigning myString to a "string" instead of String object

Upvotes: 2

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