Dmitry Olshansky , dans le message (digitalmars.D:174214), a écrit : > Most likely - since you re-read the same memory twice to do it.
You're probably right, but if you do this right after the token is generated, the memory should still be near the processor. And the operation on the first read should be very basic: just check nothing illegal appears, and check for the end of the token. The cost is not negligible, but what you do with litteral tokens can vary much, and what the lexer will propose may not be what the user want, so the user may suffer the cost of the litteral decoding (including allocation of the decoded string, the copy of the caracters, etc), that he doesn't want, or will have to re-do his own way... -- Christophe
