Author: Heiner Marxen
Date: 14:43:17 07/17/01
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On July 17, 2001 at 16:38:39, Chris Hull wrote: >On July 16, 2001 at 15:38:10, Dann Corbit wrote: > >>On July 16, 2001 at 14:58:04, Tord Romstad wrote: >> >>>On July 11, 2001 at 23:36:39, Robert Hyatt wrote: >>> >>>>2. The ANSI standards committee did the same stupid thing with bit fields that >>>>they did with other key issues, "this is left to the vendor's discretion..." >>>>Ie is a char signed or unsigned by default? Depends on the compiler. >>> >>>Please tell me you are just kidding. I have always assumed that chars are >>>signed by default, and a lot of my code depends on it. Are you saying that >>>my code could stop working when I switch to another compiler? >> >>No. Unadorned char is either signed or unsigned, depending upon the >>implementation. The reason for strange things like this is preexisting code. >>Imagine you have have a collection of hundreds of thousands of lines of code, >>and suddenly a restriction is imposed which breaks it. >> >>Personally, I would have gone ahead and broken all of the code. After all: >> >>int foo; >>long bar; >> >>do not leave the signed nature of foo and bar up to chance for whatever compiler >>is used, they must be signed > >If you don't want to leave it up to chance use > >signed int foo; >signed long bar; >unsigned int ufoo; >unsigned long ubar; > >If not you rolls the dice, you take your chances. > >Chris Hull NO, please do not! The only C type the signedness of which is not defined by the language proper is char. This is the only exception, all other (integral) types have a well defined signedness. int is always signed. To get a signed char write signed char foo; Otherwise the keyword 'signed' is useless. All of int, short and long are signed, already. Ah, and yes, there are compilers, which make plain 'char' unsigned. Regards, Heiner
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