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Home/ Questions/Q 8751485
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T13:03:24+00:00 2026-06-13T13:03:24+00:00

Why would the following be the result in ES5 non-strict mode? Object.prototype.toString.call(null); => [object

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Why would the following be the result in ES5 non-strict mode?

Object.prototype.toString.call(null);
=> [object Null]

given that

Object.prototype.toString.call(window);
=> [object global]

Shouldn’t the two lines be identical in non-strict mode, since someFunction.call(null) should be equivalent to someFunction() which should be equivalent to someFunction.call(window)?

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  1. Editorial Team
    Editorial Team
    2026-06-13T13:03:25+00:00Added an answer on June 13, 2026 at 1:03 pm

    According to ES5, when entering function code in non–strict mode:

    Else if thisArg is null or undefined, set the ThisBinding to the global object.

    so yes, it seems that this should default to the global object. But in §15.3.4.4 Function.prototype.call there is:

    The thisArg value is passed without modification as the this value. This is a change from Edition 3, where a undefined or null thisArg is replaced with the global object and ToObject is applied to all other values and that result is passed as the this value.

    Finally:

    15.2.4.2 Object.prototype.toString ()

    When the toString method is called, the following steps are taken:

    If the this value is undefined, return "[object Undefined]".
    If the this value is null, return "[object Null]".
    
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