How can templates detech if argument is let value or expression?

[ElegantBeef]

Thanks to Orc/Arc cursors and COW string literals in this example you can 
likely use the following:
    
    
    import sugar
    import strutils
    
    template test2(src:string, lab: string = "Inner tmpl") =
      let s {.cursor.} = src
      # This works for let values but fails for expressions
      let pi1 = cast[int](addr(s))
      
      let pa1 = pi1.toHex
      debugEcho lab, ": ", src, " <= ", $pa1
    
    var s1 = "Test string"
    # The following works because it is a let variable
    test2 s1, "Original s1"
    prepareMutation(s1) # This rewrites the literal so we can see cursors in 
action.
    
    let s2 = s1
    # The following also works because it is also a let variable
    test2 s2, "Copy s1"
    # The following works because it is trapped by the when branch as a literal
    test2 "Literal abc", "Literal template"
    
    test2 "--- $1" % s1, "Expression"
    
    
    Run