Re: [css-variables] Using $foo as the syntax for variables

On Wednesday, May 23, 2012 at 12:26 PM, Tab Atkins Jr. wrote:

> On Wed, May 23, 2012 at 11:11 AM, Ojan Vafai <ojan@chromium.org (mailto:ojan@chromium.org)> wrote:
> > On Wed, May 23, 2012 at 10:32 AM, Tab Atkins Jr. <jackalmage@gmail.com (mailto:jackalmage@gmail.com)>
> > wrote:
> > > The only two extensions I know I want right now are (1) the ability to
> > > specify a default value to be used when a var is invalid or undefined,
> > > and (2) the ability to grab the value of a variable from the parent
> > > (the inherited value) rather than from the current element.
> > > 
> > > Under the current syntax, these would look something like this:
> > > 
> > > color: var(bar) // normal
> > > color: var(bar, red) // default value
> > > color: parent-var(bar) // bar from the parent
> > > color: parent-var(bar, red) // bar from the parent, with a default value
> > > 
> > > In the new syntax, it would look something like this:
> > > 
> > > color: $bar //normal
> > > color: default-var($bar, red) // default value
> > > 
> > 
> > 
> > This could still be var($bar, red) and var($bar) could still work. $bar
> > could just be an alais for it. Any subset of these options for dealing with
> > default values seem fine to me.
> > 
> 
> 
> Hm, good point. No reason to use an extra-long name if a shorter one will do.
> 
This could potentially be extended further with multiple fallbacks (eg font-stacks), or with nesting functions:

margin: stack($foo, parent($bar), calc($baz + 3em), 20px);

-e

> 
> > > color: parent-var($bar) // bar from the parent
> > > color: parent-var($bar, red) // bar from the parent, with a default value
> > > 
> > 
> > 
> > The new syntax looks strictly better. It prioritizes making the common case
> > more readable (or at least more concise) and easier to use without making
> > the less common cases any worse.
> > 
> 
> 
> That's my argument, yeah.
> 
> ~TJ