Bit Operations in Spin

hylee101001

I wrote following code. But the result is little odd as commented below; it looks the data is larger than 16 bits..? Could someone explain this? Thank you.
Oh, the object I used here may not work though.. So please just put spaces at the object name and delete usb.newline. Or, use other serial object...

CON
_clkmode = xtal1 + pll16x
_xinfreq = 5_000_000

obj

usb : "ParallaxSerialTerminal.spin"

VAR

word x, one, zero, r, temp

PUB main
usb.quickstart
x := %1111_1111_0000_0101'_1111_1111_0000_0101
one := %1111_1111_1111_1111'_1111_1111_1111_1111
zero := %0000_0000_0000_0000'_0000_0000_0000_0000
temp := %0000_0000_0000_0000'_0000_0000_1111_1111

'r := x >> 8 << 8 ' outcome is as expected, 1111_1111_0000_0000
r := x << 8 >> 8 ' outcome is NOT as expected, 1111_1111_0000_0101
' I expected 0000_0000_0000_0101
repeat
usb.clear
usb.bin(r,32)
if (r>0)
usb.newline
usb.str(string("Least significant byte of x equlas 1"))
else
usb.newline
usb.str(string("Least significant byte of x NOT equlas 1"))
usb.newline


waitcnt(cnt + clkfreq/10)

Phil Pilgrim (PhiPi)

Although your variables are words, expressions are always evaluated as 32-bit signed integers.

Also,



-Phil

Duane Degn

I gave up trying to use bit operations on anything but longs a long time ago.

Some of the operations only work correctly on longs.

Ariba

If you write it like that it should result in what you expected:

...
r := x << 8
r := r >> 8
...

so the intermediate result of the first shift is written into the 16bit variable.

Andy

Dave Hein

When you do "x << 8 >> 8" the value of x is promoted to a 32-bit value before it is shifted. The result you are seeing is expected since the 16-bit value is being shifted as 32 bits.

Phil Pilgrim (PhiPi)

To get what you want, do this:

r := x << 24 >> 24

or, more simply:

r := x & $ff

-Phil

kuroneko

Phil Pilgrim (PhiPi) wrote: »

...

r := x & $ff

Alternatively one could use:

r := x.byte[0]