5 lines
285 B
OCaml
5 lines
285 B
OCaml
let f (x : < a:int; .. > as 'me1) = (x : < b:bool; .. > as 'me2);;
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let f (x : < a:int; .. > as 'me1) = (x : < a:int; b:bool; .. > as 'me2);;
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let f (x : [> `A of int] as 'me1) = (x : [> `B of bool] as 'me2);;
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let f (x : [> `A of int] as 'me1) = (x : [`A of int | `B of 'me2] as 'me2);;
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