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Home/ Questions/Q 8741691
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T11:21:42+00:00 2026-06-13T11:21:42+00:00

In Javascript, the == comparison has a strict (non-type converting) version: === . Likewise,

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In Javascript, the == comparison has a strict (non-type converting) version: ===. Likewise, != has the strict form !==. These protect you from the following craziness:

var s1 = "1",
    i1 = 1,
    i2 = 2;

(s1 == i1)   // true, type conversion
(s1 != i1)   // false, type conversion

(s1 === i1)  // false, no type conversion
(s1 !== i1)  // true, no type conversion

However, the other comparison operators have no equivalent strict modes:

(s1 < i2)   // true, type conversion
(s1 <= i2)  // true, type conversion
([] < i2)   // true, wait ... wat!?

The obvious solution seems pretty verbose:

((typeof s1 === typeof i2) && (s1 < i2))  // false

Is there a more idiomatic (or just less verbose) way to do this in Javascript?

Reference: MDN Comparison Operators

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  1. Editorial Team
    Editorial Team
    2026-06-13T11:21:43+00:00Added an answer on June 13, 2026 at 11:21 am

    There are no built-in operators for what you want, but you can always create your own functions. For example, for <:

    function lt(o1, o2) {
        return ((typeof o1 === typeof o2) && (o1 < o2));
    }
    lt("10", 11); // false
    

    Another option, if you’re only dealing with strings and numbers, is extending String.prototype and Number.prototype:

    function lt(o) {
        return ((typeof this.valueOf() === typeof o) && (this < o));
    }
    String.prototype.lt = lt;
    Number.prototype.lt = lt;
    "10".lt(11);   // false
    (11).lt("12"); // false
    
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