The grammar given in clarification E2-9 appears to be faulty. My apologies; I should have caught this when the WG reviewed it. Specifically: (1) The rule for B64x15 has too many ::= symbols. Instead of B64x15 ::= B64 B64 B64 B64 B64 ::= B64 B64 B64 B64 B64 ::= B64 B64 B64 B64 B64 I think it should read B64x15 ::= B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 B64 It's a single rule, after all, not three rules. (2) similarly for B64lastline. This also has a parenthesis missing. For B64lastline ::= B64x4? B64x4? B64x4? B64x4? ::= B64x4? B64x4? B64x4? B64x4? ::= B64x4? B64x4? B64x4? B64x4? ::= B64x4? B64x4? B64x4? B64x4? ::= B64x4? B64x4? ::= (B64x4 | B64 B64 B16 '=') | (B64 B04 '==')) ::= #xA read B64lastline ::= B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? B64x4? (B64x4 | (B64 B64 B16 '=') | (B64 B04 '==')) #xA (3) A cosmetic note: in the grammar for the lexical space, I think the rule for B64final can be better aligned. For B64final ::= B64 S? B04 S? '=' S? '=' S? | B64 S? B64 S? B16 S? '=' S? read B64final ::= B64 S? B04 S? '=' S? '=' S? | B64 S? B64 S? B16 S? '=' S? -CMSMcQReceived on Tuesday, 8 July 2003 21:12:48 GMT
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