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Hwo to read bit and set same bit in different byte


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#1 Misery   Members   -  Reputation: 317

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Posted 16 October 2012 - 05:52 AM

Hello,

I am using integers as bit containers representing bool values.
How do I make a fast instruction that reads some bit from one integer and sets its value to another.
Lets say: there are two 64 bit integers. One of them x has some bits set.
I want to rewrite every second bit from x to y (from beginning). How do I make that avoiding code similar to that:
y=0; //clear y
t=0;
for (int i=0;i<64;i+=2)
	if (x & (CONTAINER_INT(1) << (i%ContainerSize)))
   {
	  y |= CONTAINER_INT(1) << (t%ContainerSize);
	 t++;
   }

I mean I want to avoid if instruction.

Regards.

Edited by Misery, 16 October 2012 - 06:03 AM.


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#2 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 16 October 2012 - 06:10 AM

something like this?

y |= x&0x5555555555555555LL;

Edited by Olof Hedman, 16 October 2012 - 06:18 AM.


#3 Misery   Members   -  Reputation: 317

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Posted 16 October 2012 - 06:17 AM

@Olof Hedman: No. I need to rewrite them bit by bit.

#4 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 16 October 2012 - 06:25 AM

why? I assume you also do something else then, besides the copying of every second bit?

If you need to do it bit by bit, I don't think you have any choice but to check it bit by bit.

My code copies all of them from x at the same time, not caring what they are, and conserve all of the other bits from y

Edit: this if you also need to copy zeros

y = (y&0xAAAAAAAAAAAAAAAALL) | (x&0x5555555555555555LL);

Edited by Olof Hedman, 16 October 2012 - 06:31 AM.


#5 Brother Bob   Moderators   -  Reputation: 8625

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Posted 16 October 2012 - 06:35 AM

why? I assume you also do something else then, besides the copying of every second bit?

If you need to do it bit by bit, I don't think you have any choice but to check it bit by bit.

My code copies all of them from x at the same time, not caring what they are, and conserve all of the other bits from y

Your code doesn't match his. The bit index for the source and the destination are different variables; the source bit index is the loop index i, but the destination index is t which is stepped by one and only stepped for on-bits inside the if-statement. The behavior of his code is effectively: count the number of on-bits in every second bit index, and set that many bits in the destination. If the source contains 5 on-bits when counting every second bit, then the first 5 bits in the destination is set.

Although I'd say that his code and description doesn't convey the same message, and neither are very clear. A full description of the actual problem will help a lot.

#6 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 16 October 2012 - 06:42 AM

Your code doesn't match his. The bit index for the source and the destination are different variables; the source bit index is the loop index i, but the destination index is t which is stepped by one and only stepped for on-bits inside the if-statement. The behavior of his code is effectively: count the number of on-bits in every second bit index, and set that many bits in the destination. If the source contains 5 on-bits when counting every second bit, then the first 5 bits in the destination is set.

Although I'd say that his code and description doesn't convey the same message, and neither are very clear. A full description of the actual problem will help a lot.


Ah right. I actually assumed the t inside the if was a typo, since putting it outside in the for loop matches the description in the post and title

#7 Brother Bob   Moderators   -  Reputation: 8625

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Posted 16 October 2012 - 07:01 AM


Your code doesn't match his. The bit index for the source and the destination are different variables; the source bit index is the loop index i, but the destination index is t which is stepped by one and only stepped for on-bits inside the if-statement. The behavior of his code is effectively: count the number of on-bits in every second bit index, and set that many bits in the destination. If the source contains 5 on-bits when counting every second bit, then the first 5 bits in the destination is set.

Although I'd say that his code and description doesn't convey the same message, and neither are very clear. A full description of the actual problem will help a lot.


Ah right. I actually assumed the t inside the if was a typo, since putting it outside in the for loop matches the description in the post and title

Actually, putting t++ in the loop itself makes yet another interpretation of the problem: it copies bit index 2*n to bit index n, since i increases by two and t increases by one each loop in that case.

The exact details here are important and, as I said, we need to know the actual problem. So far we've had two different descriptions, and yet another interpretation of it :)

#8 Álvaro   Crossbones+   -  Reputation: 13933

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Posted 16 October 2012 - 09:21 AM

y=0; //clear y
t=0;
for (int i=0;i<64;i+=2)
	if (x & (CONTAINER_INT(1) << (i%ContainerSize)))
   {
	  y |= CONTAINER_INT(1) << (t%ContainerSize);
	 t++;
   }


y = (CONTAINER_INT(1)<<popcount(x & 0x555555555555555ul)) - 1;

EDIT: popcount stands for "population count", which means count the number of bits set. gcc provides __builtin_popcount for you, or you can use one of the tricks described in Bit Twiddling Hacks. Oh, and I am not sure why you are doing `%ContainerSize'.

Edited by alvaro, 16 October 2012 - 09:25 AM.


#9 Misery   Members   -  Reputation: 317

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Posted 16 October 2012 - 12:02 PM

OK, so little more explanation:
I have a bitset matrix (B) and two indexing integer matrices (i,j).
What I want is to create another bitset matrix that consists of B entries at the indexes defined by i and j.
It is a similar behavior like that in Matlab or Scilab.

So what I need is to rewrite B(i,j) to new bitset matrix. B entries can be 1 or 0 so I need to rewrite them bit by bit, as new matrix can have different size from B.

To complicate even more used integer depend on the platform and may have 16 32 or 64 bits. This is dependent on the template parameter CONTAINER_INT.

#10 Álvaro   Crossbones+   -  Reputation: 13933

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Posted 16 October 2012 - 12:06 PM

So your question has nothing to do with the code in your first post. :)

If I understand your problem correctly, I don't think you can do better than bit by bit.

#11 0BZEN   Crossbones+   -  Reputation: 2025

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Posted 16 October 2012 - 01:11 PM

something like this?

Not sure if the logic is correct.

y = 0;
t = 0;
while(x)
{
	y |= ((x&1) << t);
	t++;
	x >>= 2;
}

Edited by papalazaru, 16 October 2012 - 01:13 PM.

Everything is better with Metal.


#12 Álvaro   Crossbones+   -  Reputation: 13933

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Posted 16 October 2012 - 05:47 PM

You need to be much more precise than that. In your description of the problem you speak of "matrices" B, i and j, which are nowhere to be found in your code. You also don't explain what you mean by "matrix", because the conventional definition doesn't match either.

Once we have a precise statement of the problem, hopefully with some examples, I am pretty sure I'll be able to write some basic code to do what you need.

#13 Misery   Members   -  Reputation: 317

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Posted 17 October 2012 - 12:57 AM

@Alvaro:

You need to be much more precise than that. In your description of the problem you speak of "matrices" B, i and j, which are nowhere to be found in your code. You also don't explain what you mean by "matrix", because the conventional definition doesn't match either.

Once we have a precise statement of the problem, hopefully with some examples, I am pretty sure I'll be able to write some basic code to do what you need.


You're right. However, if I wanted to paste the whole code It would require about 10k lines of code at the moment, so I decided to strip it a little. And my question considers only: How to rewrite one integer to another BIT BY BIT. That is the part I am stuck at. I just have no idea how to write an instruction that does this: if the i-th bit in integer x is set, that set a j-th bit in integer y, if the i-th bit in x is unset, unset j-th bit in y. Also notice, that used integers are template parameters/arguments so no constant values should be used, rather constructors like CONTAINER_INT(1) or CONTAINER_INT(0).
Hope this clears the problem. Posted Image

PS.: And I want to do it bit by bit. I just want to avoid if statement.


EDIT: After giving it some thought, I paste the whole function (only the line with if statement has to be changed):
template <class INT_TYPE,class CONTAINER_INT>
  inline void BitsetChooseSerial(CONTAINER_INT *From,INT_TYPE FromRows,INT_TYPE *WhichRows,INT_TYPE NRows,INT_TYPE *WhichCols,INT_TYPE NCols, CONTAINER_INT *To)
  {

		 //ToRows==NRows
        static const INT_TYPE ContainerSize=sizeof(CONTAINER_INT)*8;

		 INT_TYPE ElementIndex,C,i,t,tC;

	  for (INT_TYPE j=0;j<NCols;j++)
     {

			   C=WhichCols[j]*FromRows;

			   tC=j*NRows;

		 for (i=0;i<NRows;i++)
        {

			 ElementIndex=WhichRows[i]+C;

				   t=i+tC;

		  if (From[ElementIndex/ContainerSize] & (CONTAINER_INT(1) << (ElementIndex%ContainerSize))) To[t/ContainerSize] |= CONTAINER_INT(1) << (t%ContainerSize); //change this to direct bit shift

         }
     }
    }

Edited by Misery, 17 October 2012 - 01:36 AM.


#14 Misery   Members   -  Reputation: 317

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Posted 17 October 2012 - 01:11 AM

something like this?

Not sure if the logic is correct.

y = 0;
t = 0;
while(x)
{
	y |= ((x&1) << t);
	t++;
	x >>= 2;
}


Unfortunately it doesnt work Posted Image
But thanks for help anywayPosted Image

Edited by Misery, 17 October 2012 - 01:11 AM.


#15 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 17 October 2012 - 01:18 AM

too bad shift operators are undefined with negative shifts, otherwise I'd know how :)
but if you know which index is the higher one, I think you can do something like this:

y = y&~(CONTANTER_INT(1)<<j) | (x&(CONTAINTER_INT(1)<<i)) >> (i - j))

possibly you can check which is higher outside the loop, and at least avoid the if in the loop.

#16 Misery   Members   -  Reputation: 317

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Posted 17 October 2012 - 02:02 AM

@Olof Hedman: Actually I don't get the idea of this index stuff. Why would I need to find the bigger index value?

Edited by Misery, 17 October 2012 - 02:02 AM.


#17 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 17 October 2012 - 02:12 AM

because if j > i in the above code, the shift will be negative, and negative shifts have undefined behaviour afaik

Edited by Olof Hedman, 17 October 2012 - 02:12 AM.


#18 Misery   Members   -  Reputation: 317

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Posted 17 October 2012 - 02:21 AM

@Olof Hedman: Yes, but why I would get negative shift after reading a bit value from x and placing this value at y?

#19 Olof Hedman   Crossbones+   -  Reputation: 2955

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Posted 17 October 2012 - 03:23 AM

if you want to put bit 3 from x into bit 5 of y, that is i = 3, j = 5, then you have to shift the result of x&(CONTAINTER_INT(1)<<i) to the left, instead of right as the code does.

but to write a >> -2 is not allowed.

#20 Misery   Members   -  Reputation: 317

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Posted 17 October 2012 - 03:36 AM

@Olof Hedman: Oh, now i get it. But is there no way of doing it in the flow manner like: Read bit at some i in x, write this value to some j position of y? Without tangling indexes together?




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