#11529: Show instance of Char should print literals for non-ascii printable charcters -------------------------------------+------------------------------------- Reporter: nushio | Owner: Type: bug | Status: new Priority: normal | Milestone: Component: Compiler | Version: 7.10.3 Resolution: | Keywords: Operating System: Unknown/Multiple | Architecture: | Unknown/Multiple Type of failure: None/Unknown | Test Case: Blocked By: | Blocking: Related Tickets: | Differential Rev(s): Wiki Page: | -------------------------------------+------------------------------------- Comment (by chak): Replying to [comment:2 dfeuer]:
Absolutely any code in the entire world that relies on the current behavior will break. The current behavior is expressed in the reference implementation in the Haskell 2010 report. Frankly, changing it ''is not an option''. You can write your own function to unescape valid Unicode. You can also write your own `UShow` class if you like with a method for showing various things using Unicode generally. You can then try to convince other developers to depend on your package and write instances of your class.
I disagree. I think, the current implementation is actually wrong and does not adhere to the standard. The standard states in 16.6 that `showLitChar` be defined as follows:
Convert a character to a string using only printable characters, using Haskell source-language escape conventions.
However, the current implementation of `showLitChar` fail to use `isPrint`; instead it uses a naive condition, `c > '\DEL'`, to determine printability. This is wrong. The solution is simple, replace the condition `c > '\DEL'` by `not (isPrint c)` in the definition of `showLitChar`. -- Ticket URL: http://ghc.haskell.org/trac/ghc/ticket/11529#comment:15 GHC http://www.haskell.org/ghc/ The Glasgow Haskell Compiler