| Author |
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Paminu
Guest
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 Posted:
Tue Aug 02, 2005 4:15 pm Post subject:
ALU control lines? |
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I saw an illustration of an ALU where input to Bnegate and CarryIn came from
the same signal.
Are Bnegate and CarryIn always the same? |
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Joe Pfeiffer
Guest
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| Posted:
Tue Aug 02, 2005 4:15 pm Post subject:
Re: ALU control lines? |
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Paminu <jadajada@asd.com> writes:
| Quote: | I saw an illustration of an ALU where input to Bnegate and CarryIn came from
the same signal.
Are Bnegate and CarryIn always the same?
|
Why do you suppose they are the same in the illustration you're
looking at? Does that help you to see whether they should always be
the same?
--
Joseph J. Pfeiffer, Jr., Ph.D. Phone -- (505) 646-1605
Department of Computer Science FAX -- (505) 646-1002
New Mexico State University http://www.cs.nmsu.edu/~pfeiffer
skype: jjpfeifferjr |
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Guest
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| Posted:
Tue Aug 02, 2005 4:16 pm Post subject:
Re: ALU control lines? |
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If all you want to do is Add and Subtract.....
a - b = a + negate(b) + 1 |
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Paminu
Guest
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| Posted:
Wed Aug 03, 2005 12:15 am Post subject:
Re: ALU control lines? |
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rhnlogic wrote:
| Quote: | Paminu wrote:
MitchAlsup@aol.com wrote:
If all you want to do is Add and Subtract.....
a - b = a + negate(b) + 1
Note the above equation is for a subtract without either a borrow
or a reverse annuity with per clock cycle compound interest.
I have made a 1-bit ALU where Binvert and cIn get the same signal (from
Bnegate). But the result is incorrect when I would like to make the
following addition:
a = 0
b = 0
cIn = 1
Why did you choose to call this input "cIn" , instead of "Binvert"
or something else even more appropriate?
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There are both cIn (carryIn) and binvert. Point is that Bnegate goes to both
cIn and binvert. If Bnegate is 1 then cIn and binvert is also 1. But when
binvert is 1, negate(b) is choosen and this results in:
0+1+1= 10
instead of:
0+0+1 = 1 |
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rhnlogic
Guest
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| Posted:
Wed Aug 03, 2005 12:15 am Post subject:
Re: ALU control lines? |
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Paminu wrote:
| Quote: | MitchAlsup@aol.com wrote:
If all you want to do is Add and Subtract.....
a - b = a + negate(b) + 1
|
Note the above equation is for a subtract without either a borrow
or a reverse annuity with per clock cycle compound interest.
| Quote: | I have made a 1-bit ALU where Binvert and cIn get the same signal (from
Bnegate). But the result is incorrect when I would like to make the
following addition:
a = 0
b = 0
cIn = 1
|
Why did you choose to call this input "cIn" , instead of "Binvert"
or something else even more appropriate?
IMHO. YMMV.
--
rhn |
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Paminu
Guest
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| Posted:
Wed Aug 03, 2005 12:15 am Post subject:
Re: ALU control lines? |
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Joe Pfeiffer wrote:
| Quote: | Paminu <jadajada@asd.com> writes:
I saw an illustration of an ALU where input to Bnegate and CarryIn came
from the same signal.
Are Bnegate and CarryIn always the same?
Why do you suppose they are the same in the illustration you're
looking at? Does that help you to see whether they should always be
the same?
|
If a=0, b=0 and cIn=1 the result will be "10", since binvert will also be 1.
But 0+0+1 should be "1" and not "10" so I don't understand why the two
inputs gets the same signal. |
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Paminu
Guest
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| Posted:
Wed Aug 03, 2005 12:15 am Post subject:
Re: ALU control lines? |
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MitchAlsup@aol.com wrote:
| Quote: | If all you want to do is Add and Subtract.....
a - b = a + negate(b) + 1
|
I have made a 1-bit ALU where Binvert and cIn get the same signal (from
Bnegate). But the result is incorrect when I would like to make the
following addition:
a = 0
b = 0
cIn = 1 (this implies that binvert will also be 1 and therfore negate(b)
will be chosen).
I then get 0(a) + 1(negate(b)) + 1(cIn) = 10
where 0 is the sum and 1 is cOut.
If I seperate the signals and set binvert to 0, I get:
0(a) + 0(b) + 1(cIn) = 1
where 1 is the sum and cOut will be 0.
So I guess that even if I want to make a simple addition I need to seperate
the signals. |
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