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Gradient helper prevents having a 'top()' mixin #56

@ianstormtaylor

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@ianstormtaylor

I get this error when trying to declare a mixing with a name of top()

TypeError: expected string, ident or literal, but got function:top()

and declaring a gradient using a string doesn't work either linear-gradient('top', ...

Not sure if there is a solution that keeps both of the functions around...

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