Hi,
i want a sketch that would return the nearest multiple of 8 (rounded up always) from a variable, for example.
X = 4, returns 8
X=1, returns 8
X=8, returns 8
etc...
X=11, returns 16
X=14, returns 16
etc...
X=20, returns 24
X=17, returns 24
etc
How so?
Say X = 4,
4/8 = 0.5
0.5 + 1 = 1.5
1.5*8 = 12 (Should be 8 ).
edit: Whoops, i dont know what im on about here, should be 8
Something like
x = (x + 7) & ~7 ?
4/8 = 0.5
No. 4/8 = 0
4.0/8.0 = 0.5. Not the same thing at all.
adilmalik:
How so?Say X = 4,
4/8 = 0.5
0.5 + 1 = 1.5
1.5*8 = 12 (Should be 16).
x = 4
4 / 8 = 0
0 + 1 = 1
1 * 8 = 8
Oh i see, Thankyou!, ive come up with this, dont have an arduino at hand to test it, what do you guys think:
X = 8 * ceil((X / 8))
I think ceil() takes a float argument, so if x is an int then use ceil(x/8.0), because ceil(x/8) will always round down
if x is an int, I believe the modulus operator is faster than a float division
if(!(x%8))answer=x;
else answer=x-(x%8)+1;
or the shorthand version
answer=(x%8)?x-(x%8)+1:x;
or ... (I think)
answer = ((x >> 3) + 1) << 3);
...R
Robin2:
or ... (I think)answer = ((x >> 3) + 1) << 3);...R
try x== 0
I had the same thought 
the addition of 1 should be conditional something like
answer = ((x >> 3) + ((x&7)>0) << 3);
//assumes true == 1 and false == 0
x = (x + (n-1)) & ~(n-1);
this only works for powers of two, due to the sexy nature of how bit masking works - in all other cases - you do this:
x = ((x + (n-1)) / n) * n); // edited
so, to answer the fundamental question here - AWOL was spot on here with the solution, unfortunately since it was low level bitwise C it seems that many have failed to understand what was actually going on.
x = (x + 7) & ~7;
x = ((x + 7) / 8) * 8); // edited
do exactly the same thing.. push x over the 8 boundary (but, only by 7) and then divide ..
x = (x + 7) & ~7;
x = (x + 7) / 8;
do exactly the same thing
No, they don't.
The first one is correct (for positive integers). The second one isn't even close.
Pete
el_supremo:
No, they don't.
The first one is correct (for positive integers). The second one isn't even close.
woops 
forgot the multiply - need more coffee.
x = ((x + 7) / 8) * 8;
i would have used the first method (as reported by AWOL) anyhow 
Heh - I'm just downing a cup of tea and a chocolate bar.
Yup - AWOL's solution is definitely the way to it.
Pete
robtillaart:
try x== 0the addition of 1 should be conditional something like
I admit I had not thought of that.
But we don't expect division by 0 to work so we check for it before we try the division - same here. perhaps?
Rather than make the formula more complex?
...R
x = 0;
x = (x + 7) & ~7;
gives the result of x = 0 - there is no need to have a conditional for it.
ardiri:
x = 0;
x = (x + 7) & ~7;
gives the result of x = 0 - there is no need to have a conditional for it.
I think it works, but would you care to explain the maths?
...R
Robin2:
I think it works, but would you care to explain the maths?
we know it works
are you familiar with binary (0's and 1's) and how integers are stored internally?
anyhow, here is the catch - since it is a multiple of a power of 2, we can use the magic trickery of bit wise operations. & ~7 really means - remove the lower three bits from the number (7 = 111 in binary, ~7 = 111111...111111000 in binary) - which really means make them 0. so if you have a number like 181 it is in binary as: 10110101 making the last three bits 0 would make the number: 10110000 which would be the number 176. you can read up more at:
Bitwise operation - Wikipedia <-- better explanation
as i mentioned earlier, when the number (n) isn't a power of 2, you need to use another formula:
x = ((x + (n-1)) / n) * n);
as the bit masking technique wont work then.
ardiri:
we know it worksare you familiar with binary (0's and 1's) and how integers are stored internally?
I had figured out the stuff in your post. What I am interested to understand is why it works as maths rather than just as bit manipulations. The bit manipulation stuff is HOW not WHY.
...R