javascript中的单引号和双引号值比较 - 严格类型

Question

From the below code when I want to compare variable y and z, here strict type of comparison is matching, but y==z is not matching. Can anybody explain the reason and how strict type works ?

<script language="javascript" type="text/javascript">
    var x=11;
    var y="11";
    var z='11';

    if(x==y)
        alert('x==y');
    if(y==z)
        alert('y==z');
    if(x===y)
        alert('x===y');
    if(y===z)
        alert('y===z');
    if(x==z)
        alert('x==z');
    if(x===z)
        alert('x===z');
</script>

Answer 1

Both are string and also it's matching.

> typeof "11";
"string"
> typeof '11';
"string"
> "11" === '11'
true
> "11" == '11'
true

Answer 2

JavaScript has both strict and type-converting equality comparison. For strict equality the objects being compared must have the same type and:

Two strings are strictly equal when they have the same sequence of characters, same length, and same characters in corresponding positions. Two numbers are strictly equal when they are numerically equal (have the same number value). NaN is not equal to anything, including NaN. Positive and negative zeros are equal to one another. Two Boolean operands are strictly equal if both are true or both are false. Two objects are strictly equal if they refer to the same Object. Null and Undefined types are == (but not ===). [Ie (Null==Undefined) is true but (Null===Undefined) is false]

Check this out

Answer 3

When you use double equals ( Loose equality ) you only compare values , but when using three equals ( Strict equality ) you compare types too.

Loose equality compares two values for equality, after converting both values to a common type.

So, if

var y = 11; // number
var z = '11'; // string

var equalValues = y == z; // true
var equalValuesAndTypes = y === z; // false

You can check the MDN docs