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Subject: Re: Outputting u64 word using printf?

Author: Robert Hyatt

Date: 10:58:18 06/07/04

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On June 07, 2004 at 13:13:33, Vincent Diepeveen wrote:

>On June 07, 2004 at 10:40:09, Fabien Letouzey wrote:
>
>>On June 07, 2004 at 10:36:58, Vincent Diepeveen wrote:
>>
>>>On June 07, 2004 at 10:30:50, Fabien Letouzey wrote:
>>
>>>r = 1<<36; // works for good compilers not for the rest
>>
>>Hm, my understanding is that first "1<<36" should be calculated (so it's 0 if
>>sizeof(int) <= 4) and *then* cast into r's type (oops, too late).
>>Fabien.
>
>The result is undefined when integer = 32 bits and long long is 64 bits.
>
>Trivially it is a very natural thing to try to write r = 1 << 36 when using a 64
>bits type;
>
>Needing to cast every non-integer is not a nice way of programming now that all
>processors are getting 64 bits, and 99% of all programmers will make mistakes
>with it, basically taking care C/C++ in the future is not a serious programming
>language.
>
>Because ansi-C definitely runs behind here on good programming habits, it's up
>to the compilers mercy whether it works or doesn't, so only good compilers will
>support it, as they support a natural way of working with the environment.
>
>The rule in ansi-C should read that for constants assigned, they should be
>assumed to have the same type and not that they should be assumed to be integer.


That won't work.  i = .1 * 20.0; would produce errors if you convert .1 and 20.0
to integers first, then do the multiply.

_NO_ compiler will think that 1 << 36 is anything but zero on an architecture
where int == 32 bits.  Making them do so would break more programs than it would
help.

The ANSI C standard certainly has plenty of holes. But this really isn't one of
them that is a problem...



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