Hello
If i set the input of my block as a byte, means that the block will
handle 8
bits at a time and each item correspond to one bit of that byte?
in[0] is the first bit and so on?
Thank you.
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On Tue, Jun 16, 2015 at 2:17 PM, Marbellys
[email protected]
wrote:
If i set the input of my block as a byte, means that the block will handle
8
bits at a time and each item correspond to one bit of that byte?
in[0] is the first bit and so on?
No, this isn’t automatic. You would also have to precede your block
with a
block like gr::blocks::unpack_k_bits_bb or ::repack_k_bits_bb to
generate
such a stream from a packed byte format.
Thank you, Johnathan.
So if i use an unpack_k_bits block, and i set k to 1, will I be handle
1
bit at a time?
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On Tue, Jun 16, 2015 at 4:13 PM, Marbellys
[email protected]
wrote:
So if i use an unpack_k_bits block, and i set k to 1, will I be handle 1
bit at a time?
Yes.
Dear Marbelly,
unpack_k_bits with k = 1 means, for instance,
a byte sample 0x0F (0b0000 1111) in a stream
is unpacked to 0x00 0x00 0x00 0x00 0x01 0x01 0x01 0x01 (MSB first)
and only the first insignificant sample is taken and the rest 7 samples
are
discarded.
So if you want to process 8 bits in a byte individually, use
unpack_k_bits
with k = 8.
Regards,
Jeon.
2015-06-17 11:49 GMT+09:00 Marbellys [email protected]:
Hi Jeon.
Thank you very much for your response.
So if I have for example a vector source V =
(1,1,0,1,0,0,1,0,0,0,1,0,1,1,0,1 (which output’s type is set to byte),
the
first stream at the output will be (1,1,0,1,0,0,1,0), is this right?
Then, if I use an unpack_k_bits with k = 8 at its output, will I be able
to
save each of this bits in a vector in a new block and to manipulated
them
individually?
Thank you.
Marbellys.
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