I find it misleading that if I create a form element using Javascript with a particular value attribute, and then use Javascript to alter the value attribute that Firebug shows only the original value, and not the new value. For example imagine I have the following HTML:
<p id='row'>
<input type='text' id='Fred' name='Fred' value='Flintstone'>
</p>
And in Javascript I do:
var newinput = document.getElementById('Fred').clone(false);
newinput.id='Barney';
newinput.name='Barney';
newinput.value='Rubble';
document.getElementById('row').append(newinput);
Then Firebug displays the value attribute of the new input node as "Flintstone", not "Rubble", while it shows the id and name attributes as 'Barney'.
This is the situation where element.value is not the same as element.defaultValue. Because of the clone operation newinput.defaultValue is "Flintstone" and newinput.value is "Rubble". To make the Firebug display of the input element display the current value of the value attribute, I must also change the value of the defaultValue attribute so the two attributes agree. I feel that in this situation Firebug should either display just the current value of the value attribute, which is what the external user sees in the browser, or should display the values of both the value attribute and the defaultValue attribute.
This is with Firebug 2.0.6 on Firefox 34.0 running on Ubuntu Linux.