Re: [Dev-C++] Shift operators portability

[Per W. <pw...@ia...>]

Since we are talking about << and >> for shifting bits, I still assume
we are talkinga bout integer arithmetic and not floating point. In that
case * and / won't involve any fp unit.

Anyway - the shift operators are very much wanted now too. It is simpler
to use them that to try and fiddle around with mult/div when you work with
bit manipulation. Yes, it is possible to remove | & ~ ^ << and >> but
manipulation of bit flags wouldn't directly become simpler.

// with bit manipulation
void process_alarms(unsigned alarms) {
    unsigned flag = 1;
    while (alarms) {
        if (alarms & flag) {
            cout << "processing alarm " << flag << "\n";
            alarms &= ~flag;
        }
        flag <<= 1;
    }
}

// without bit manipulation
void process_alarms(unsigned alarms) {
    unsigned mask = 2;
    while (alarms) {
        mod = alarms % mask;
        if (mod) {
            cout << "processing alarm " << mod << "\n";
            alarms -= mod;
        }
        mask *= 2;
    }
}

/Per W

On Tue, 23 Sep 2003, Michael D. Smith wrote:

>
> OK, with that "slight*" correction. It does multiply and divide by 4 (using
> a 2), and as long as it's small numbers, and not negative numbers, it
> appears to work reliably both ways and back and forth.
>
> 30 or 40 years ago, when computers were a lot slower and had almost no
> memory, using this might have made sense. With a gigahertz computer, you're
> probably much better off to just stick with the regular math functions. If
> they are any slower, it's not by any noticeable amount, they are definitely
> more portable, and much less error prone -- programmer, maintainer, and
> otherwise (the dropped bits).
>
> I'm not real sure that when you engage the math processor to multiply or
> divide by 2,4,8,16 ect, that it doesn't just do this anyway -- only in a
> more portable/reliable fashion.
>
> Bottom line what's the point???
>
> It is fun to play with though. Think about being able to control one little
> bit in hundreds of millions of bytes -- now all I gotta do is figure out
> which one, and what I'm gonna do with the rest of 'em :)
>
> ms
>
>
> * just the difference between working and not working
>
>
>
> No wraparound, they drop off the end, at least on any machine I've ever
> been familiar with.
>
> I thought I would try, see how it works on this machine and got nothing. No
> compiler error, and no change in the int.
>
> Anyone know why this produces nothing (16) every time? And I tried other
> numbers, in both positions, as well. Even tried turning it around -- Same
> thing, nothing, every time. It seems there should be some effect, or an error..
>
> I have an AMD processor and no idea if it's big or little endian.
>
>
>
> #include<iostream.h>
> #include<stdlib.h>
> #include<fstream.h>
>
>    int main()
>    {
>       int count = 16;
>
>       count = count >> 2;
>         cout << "count = " << count << "\n";
>
>       count = count << 2;
>         cout << "count = " << count << "\n";
>
>       system ("pause");
>       return 0;
>    }
>
>
> At 08:14 PM 9/22/03, you wrote:
> >Hi there.
> >
> >Are the shift operators ( << and >> ) portable?
> >i.e.
> >Does x << 1 have the same effect as multiplying by 2 ?
> >and
> >Does x >> 1 have the same effect as dividing by 2?
> >On both big and little endian order machines?
> >
> >also is there any wraparound?
> >i.e. in a shift the bits that are pushed out of the data structure placed
> >at the back-end of the data structure (on the other side of where they
> >were pushed out)?
> >
> >Thanks in Advance
> >David McKen
> >
> >
> >Do you Yahoo!?
> ><http://us.rd.yahoo.com/evt=10469/*http://sitebuilder.yahoo.com>Yahoo!
> >SiteBuilder - Free, easy-to-use web site design software
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>
>
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