Decode value |< "value"

MAElektronik

A simple question using bitwise decode operator.

Example code:
If |<value & % 0000_0001
"Do something"
else
"Do something else"

What happens if value is outside 0 to 31 ?
What if value == TRUE (-1 decimal) ?

Wuerfel_21

I'm decently sure it just wraps around. I.e. |<x == |<(x&31). So |<(-1) == |<31 == $8000_0000.

Note that If |<value & %0000_0001 could just be written as ifnot value&31, but I guess that's just your example nonsense.

JonnyMac

Ada is correct. Think of the |< x as 1 << (x & $1F) A bit of test code proves that.

Wuerfel_21

@JonnyMac said:
Ada is correct. Think of the |< x as 1 << (x // 32). A bit of test code proves that.

Well, really just 1 << x, since the shift operators have the same wraparound behaviour

JonnyMac

I just noticed that, which was was a surprise. I had always that believed that shifting with a value > 31 resulted in 0.

Phil Pilgrim (PhiPi)

As I understand it, the Propeller uses a "barrel shifter", so nothing really get "shifted." It's just a fancy set of multiplexers that take part of the input and gate it somewhere else into the result. The multiplexers themselves are gated by a decoder, whose inputs are the lower five bits of the shift amount.

-Phil

MAElektronik

Thanks everybody 😊

Perhaps the manual should have added this information?