Odd JavasScript string replace behavior with $&

Question

With the following code:

var x = 'foo';
console.log(x.replace(x, "\\$&"));​

The output is '\foo', as shown here: http://jsfiddle.net/mPKEx/

Why isn't it

'\\$&"?

I am replacing all of x with "\$&" which is just a plan old string so why is string.replace doing some crazy substitution when the 2nd arg of the function isn't supposed to do anything except get substitued in...

Answer 1

$& is a special reference in Javascript's string replace. It points to the matched string.

$$ - Inserts a "$"
$& - Refers to the entire text of the current pattern match. 
$` - Refers to the text to the left of the current pattern match. 
$' - Refers to the text to the right of the current pattern match.
$n or $nn - Where n or nn are decimal digits, inserts the nth parenthesized
            submatch string, provided the first argument was a RegExp object.

(Reference)

Answer 2

In your case:

var x = 'foo';
console.log(x.replace(x, function() {return '\\$&'}));

See the differences: http://jsfiddle.net/mPKEx/10/

You can specify a function as the second parameter. The above-mentioned special replacement patterns ($$, $&, $`, $', $n or $nn) do not apply in this case.