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

[stefan k. <ste...@ya...>]

Something to keep in mind about this matter:

You probably do not have to explicitly use the >> or << operators, since 
most modern compilers will optimize the code to use shifts where 
possible anyway.

Of course that only works, if the divisor or multiplicator is a constant 
and the division / multiplication can be performed by using a single 
shift instruction. (i.e. 2/4/8/16/...)

So instead of writing:

const long kFactor = 2;

value = foo << kFactor;

You can as well simply write:

const long kFactor = 4;

value = foo * kFactor;

Which will result in a left-shift by 2, effectively. (With most C/C++ 
compilers today.)

The *big advantage* of this is, that you can later easily change 
'kFactor' to three 3 (if that should become necessary), without having 
to find and change all your shifts back to multiplications in your code!

Cheers,
Stefan

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! 
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> 
> 
> 
> 
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