Javascript "Not a function"

Question

The below given code gives an error arguments.sort is not a function. Is it because arguments object cannot be altered directly? Or is it something else.

Any help would be appreciated.

function highest()
{ 
    return arguments.sort(function(a,b){ 
         return b - a; 
    }); 
} 
assert(highest(1, 1, 2, 3)[0] == 3, "Get the highest value."); 
assert(highest(3, 1, 2, 3, 4, 5)[1] == 4, "Verify the results.");

The assert function is as follows (Just in case)

function assert(pass, msg){
   var type = pass ? "PASS" : "FAIL";
   jQuery("#results").append("<li class='" + type + "'><b>" + type + "</b> " + msg + "</li>");
}

elclanrs · Accepted Answer · 2013-03-07 09:53:44Z

4

Try this:

return [].sort.call(arguments, function(a, b) {
   return b - a;
})

Edit: As @Esailija pointed out, this doesn't return a real array, it just returns the arguments object which is an array-like object. It's fine for iterating and accessing properties by index, but that's about it.

edited Mar 7, 2013 at 9:53

answered Mar 7, 2013 at 9:40

elclanrs

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2 Comments

Esailija Over a year ago

But this doesn't return an array :x

elclanrs Over a year ago

@Esailija: Right, didn't notice that, it returns an arguments object. It should be fine for this case, but yeah OP might want an actual array.

Esailija · Accepted Answer · 2013-03-07 09:40:25Z

2

It is because arguments is not an array and doesn't have the sort method.

You can use this trick to convert it to an array:

function highest()
{ 
    return [].slice.call(arguments).sort(function(a,b){ 
         return b - a; 
    }); 
}

answered Mar 7, 2013 at 9:40

Esailija

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Comments

Alex Pereira · Accepted Answer · 2013-03-07 09:47:09Z

0

You highest function isn't passing any argument inside

function highest(arguments)
{ 
    return arguments.sort(function(a,b){ 
         return b - a; 
    }); 
}

and should work with an array

highest([1, 1, 2, 3])[0]

answered Mar 7, 2013 at 9:47

Alex Pereira

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