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Home/ Questions/Q 6130045
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T16:48:45+00:00 2026-05-23T16:48:45+00:00

I am curious as to why this works: var c = { d: function

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I am curious as to why this works:

var c = {
    d: function myFunc() {
        console.log(this === window);   
    }
};
var a = {
    b: function() {
        console.log(this === a);
        (0,c.d)();
        c.d();
    }
};
a.b();

Console output:

True
True
False

So it seems to be that (0, c.d)() is the same as c.d.call(window), but I cannot seem to find much about why or how this works. Can anyone explain?

From: Closure Compiler Issues

Fiddle: http://jsfiddle.net/wPWb4/2/

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  1. Editorial Team
    Editorial Team
    2026-05-23T16:48:45+00:00Added an answer on May 23, 2026 at 4:48 pm

    If you write multiple expressions separated by a comma (,), then all expressions will be evaluated, but you will end up with the value of the last expression:

    var x = (1,2,3);
    console.log(x); // this will log "3"
    

    Now (0,c.d) is an expression that will return the function c.d, but now c is not the this of the function anymore. This means that this will point to the global object (window), or remain undefined in strict mode. You would get the same effect with any of these:

    var f = function(x) { return x; };
    f(c.d)();
    

    Or just

    var f = c.d;
    f();
    
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