Pin value to binary number

Archiver

You may use direct bit to bit correspondece like:
x.bit0 = in0
x.bit1 = in1
x.bit2 = in2
or by nibble:
x var nib
x= inA 'low nible of low byte
or by byte
x var byte
x= inL
or by word
x var word
x =inS

ACJacques

rgolub@l... wrote:
>
> I think this is an easy question but I can't figure it out:
>
> Reading pin states:
>
> Value1 = IN1
> Value2 = IN2
> Value3 = IN3
>
> what I want to do is to create a binary value that represents the
> values of the pins, ie. if Value1 = 1, Value2 = 0, Value3 = 1 then
> the result would be (binary) 101.
>
> How does one do that?
>
> Thanks to all....
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

Archiver

I think this is an easy question but I can't figure it out:

Reading pin states:

Value1 = IN1
Value2 = IN2
Value3 = IN3

what I want to do is to create a binary value that represents the
values of the pins, ie. if Value1 = 1, Value2 = 0, Value3 = 1 then
the result would be (binary) 101.

How does one do that?

Thanks to all....

Archiver

I guess there is a number of ways to do this
1 if you can use one var rather than three

value var nib
value = ina 'if ina = 1101
value <<1 >>1 'this just drops the 4th bit

now
value.bit0 = 1
VALUE.BIT1 = 0
VALUE.BIT2 = 1



rgolub@l... wrote:
>
> I think this is an easy question but I can't figure it out:
>
> Reading pin states:
>
> Value1 = IN1
> Value2 = IN2
> Value3 = IN3
>
> what I want to do is to create a binary value that represents the
> values of the pins, ie. if Value1 = 1, Value2 = 0, Value3 = 1 then
> the result would be (binary) 101.
>
> How does one do that?
>
> Thanks to all....
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

Archiver

or in your case you can shift right 1 this will make in1 = in0 2 will be
1 and 3 will be 2

another way is

value.bit0 = in1
value.bit1 = in2
value.bit2 = in3

now the nib varable value = X101, x is not used so you may want to shift
left one then back to make the fourth bit a 0.

"L .Gaminde" wrote:
>
> I guess there is a number of ways to do this
> 1 if you can use one var rather than three
>
> value var nib
> value = ina 'if ina = 1101
> value <<1 >>1 'this just drops the 4th bit
>
> now
> value.bit0 = 1
> VALUE.BIT1 = 0
> VALUE.BIT2 = 1
>
>
>
> rgolub@l... wrote:
> >
> > I think this is an easy question but I can't figure it out:
> >
> > Reading pin states:
> >
> > Value1 = IN1
> > Value2 = IN2
> > Value3 = IN3
> >
> > what I want to do is to create a binary value that represents the
> > values of the pins, ie. if Value1 = 1, Value2 = 0, Value3 = 1 then
> > the result would be (binary) 101.
> >
> > How does one do that?
> >
> > Thanks to all....
> >
> >
> >
> > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

Archiver

--- In basicstamps@y..., "L .Gaminde" <lgaminde@t...> wrote:
> or in your case you can shift right 1 this will make in1 = in0 2
will be
> 1 and 3 will be 2
>
> another way is
>
> value.bit0 = in1
> value.bit1 = in2
> value.bit2 = in3
>
>
That works! Thanks.

Archiver

rgolub -

While the answers given are fine, the Stamp provides
another approach that can make for fewer operations in
your code, make the code more understandable, and give
you exactly what you asked.

Let's say that you use IN1, IN2, and IN3, and tie an
unused IN0 to ground (so that it remains at logic 0).
These four input bits are combined automatically as
INA (see p. 47 of the BASIC Stamp Programming Manual
2.0c, or page 216 of BASIC Stamp Manual 1.9).

INA = 8*IN3 + 4*IN2 + 2*IN1 + IN0

So that instead of:

IF IN1 = 1 AND IN3 = 1 THEN SubRoutineX

you could write:

IF INA = 10 THEN SubroutineX

Not only does the command call for less logic
operations, but it makes clear anything you missed.
There are no odd values for INA in this case, because
IN0 is tied to ground. But what if IN2 also is high?
The branch to SubroutineX would not take place,
because the value of INA would be 14. In a controls
situation, that's a good thing as the event of all
three inputs being high might otherwise be unforeseen.


I have a project currently where I am checking to see
which of four bits is set to determine which
operational mode to go to. Consider the code:

ChooseMode:
IF IN0 = 1 THEN Mode1Subroutine
IF IN1 = 1 THEN Mode2Subroutine
IF IN2 = 1 THEN Mode3Subroutine
IF IN3 = 1 THEN Mode4Subroutine
GOTO ChooseMode:

What if I switch from Mode 2 to Mode 3? If I have a
non-shorting switch (break-then-make), fine; I see
logical 0 on all four bits and loop back until the
switch energizes the third position and I see a
logical 1 on IN2 so that I select the third mode. But
if I have a shorting switch (make-then-break), then
momentarily both IN1 (for Mode 2) and IN2 (for Mode 3)
are on, but since I get to the IF statement for Mode 2
first, I go back to Mode 2, when I wanted to go to 3!
With the named nibble, I can avoid this problem:

ChooseMode2:
IF INA = 1 THEN Mode1Subroutine
IF INA = 2 THEN Mode2Subroutine
IF INA = 4 THEN Mode3Subroutine
IF INA = 8 THEN Mode4Subroutine
GOTO ChooseMode2:

Now, not only do I not have to worry about shorting
versus non-shorting switches, but if the switch
becomes defective and stays on two modes, I get a
value that is the aggrgate of two bits such as 3, 6,
or 12, but I safely avoid misrouting control.

Bob Pence



--- rgolub@l... wrote:
> I think this is an easy question but I can't figure
> it out:
>
> Reading pin states:
>
> Value1 = IN1
> Value2 = IN2
> Value3 = IN3
>
> what I want to do is to create a binary value that
> represents the
> values of the pins, ie. if Value1 = 1, Value2 = 0,
> Value3 = 1 then
> the result would be (binary) 101.
>
> How does one do that?
>
> Thanks to all....
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
>


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