A couple of days ago I posted a couple of initialisers and four operators that resolve the ergonomic issues I personally was looking to see solved in working with buffers of ints. I’ve prepared a patch and this change passes all the Swift project tests without requiring the new literal syntax. As these operators are targeted, this new approach does do not allow nonsense expressions such as '1' / ‘1’ or 'x'.isMultipleOf('a’) as was the case when we were exploring using the ExpressibleBy protocols to make Character literals convertible to integers which seems to have been a misstep.
This nicely decouples the decision on whether Swift needs a new literal syntax from our ergonomic goals and we can make a decision about it separately. For me it is still worth pursuing as people have slipped into the new syntax in posts indicating it seems very familiar and a distinction worth making. This will be a source breaking change (why introduce a new syntax without deprecating the old one) for which the bar has to be set high. To give an idea of the amount of disruption, this would require changes in 96 places in all of the standard library and 58 changes in the Swift project tests. We seem to be headed towards Unicode.Scalar literals and not Character literals this would be a retrograde step IMO. Character is the element type of Swift strings and what constitutes a single visual entity on the screen. Character is not a particularly useful type but subtle distinction between Unicode.Scalar and Character is not one we want users to have to be mindful of to use the new literal.
Looking back through the posts it seems at times we took the fork to Unicode.Scalar literals to facilitate implementing the trapping .ascii property. This is not a consideration with the patch I propose as this property is not required. This implicit choice of the encoding may be viewed as a mis-feature by some but for me ASCII is so pervasive and would be most peoples understanding (particularly with the new literal syntax) encountering an expression such as:
if cString.advanced(by: 2).pointee == ‘i' {