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Hi,
Can someone give me the differences between & and % ?
I have the following algorithm which limits a random variable to 255 only.
int nRand = rand();
nRand %= 255;
Now , doesn't the last statement have similar effect when replacing it with this.
nRand &= 255;
?
What do those operations do?Quote:
Originally Posted by Swoorup
Actually, that would set 254 to be the upper limit, but it presumably is a typo and you wanted to write 256.Quote:
Originally Posted by Swoorup
Yes (once the typo is corrected), but that is just a coincidence.Quote:
Originally Posted by Swoorup
So, you mean, they are the same
if I change nRand %= 255 to
nRand %= 256?
No, they are not the same. They will just have the same effect of limiting the range of the value.Quote:
Originally Posted by Swoorup
I see, thank you.
Which one would be the most effective for picking random though?
Changing x %= 256 into x &= 256 is an optimisation many compilers can do for themselves.
If you're going to optimise anything, then focus on the whole algorithm (like bubble sort vs. quicksort) where your skills can make a real difference.
Micro optimisations at the single operator level won't do much for you.
Hmm, thank you
Is the opmization of the compiler like this though? when x %= 256 then x &= 255 not x &= 256 :p
I am learning about the lower-level details of C++ and finding out the relation between assembly and C++ via debugging them. So, minor things like this happened to boggle my mind :P
Maybe. It depends on whether the compiler can prove that the result will always be the same. For example, -1 % 256 == -1, but -1 % 255 == 255, assuming two's complement. Now, we know for sure that rand() returns a non-negative integer, but the compiler may or may not be able to determine this, or if it can, it might not have done so. Then, there's the issue of proving that for every non-negative integer i in the range of int, (i % 256) == (i & 255).Quote:
Originally Posted by Swoorup
Seems to be true ....