How can I do a circular bit shift in propeller C

Invent-O-Doc · 2013-10-13 18:13

I know people use statements all the time like a = z << 2; , which gives a bit shift and everything that goes to the end of the variable's bit range gets chopped off, but doesn't the propeller have a PASM for a bit rotation that takes bits from one side to another in the form of a rotation???

SPIN has the '->' and '<-' operators that do this.

Is there any equivalent to this rotation command using C or C++ in Simple IDE????

The examples I see on C sites are fairly inefficient compared to an equivalent of the prop's own bit rotate.

kuroneko · 2013-10-13 18:43

I wouldn't worry too much. The compilers are quite clever in spotting the intent, i.e. propgcc will convert

(a << b)|(a >> (32 - b)) to [COLOR="#020FC0"]rol a, b[/COLOR] and
(a << (32 - b))|(a >> b) to [COLOR="#FF8C00"]ror a, b[/COLOR]

Heater. · 2013-10-14 02:17

kuroneko,

The compilers are quite clever in spotting the intent

Not in this case with propgcc it seems:

#include <stdio.h>
#include <math.h>

int roller (int a, int b)
{
        return ((a << b)|(a >> (32 - b)));        // To rol a, b and
}

int main (int argc, char* argv[])
{
        return (roller(3, 9));
}

Compiles down to:

.text
        .balign 4
        .global _roller
_roller
        mov     r7, #32
        sub     r7, r1
        mov     r6, r0
        sar     r6, r7:Q
        shl     r0, r1
        mov     r7, r6
        or      r7, r0
        mov     r0, r7
        lret
        .balign 4
        .global _main
_main
        mvi     r0,#1536
        lret

Using: /opt/parallax/bin/propeller-elf-gcc -Os -Wall -S -o rol.s rol.c

As we see there are two shifts and an OR operation.

Similar with GCC on my PC.

Don't we have an intrinsic in propgcc to do this inline?

kuroneko · 2013-10-14 02:30

Fair enough, I used unsigned int:

unsigned int rol(unsigned int a, int b) {
  return (a << b)|(a >> (32 - b));
}

unsigned int ror(unsigned int a, int b) {
  return (a << (32 - b))|(a >> b);
}

int main()                                    // Main function
{
    return rol(3, 9);
}

with your command line I get:

.text
	.balign	4
	.global	_rol
_rol
	rol	r0, r1
	mov	pc,lr
	.balign	4
	.global	_ror
_ror
	ror	r0, r1
	mov	pc,lr
	.balign	4
	.global	_main
_main
	mvi	r0,#1536
	mov	pc,lr

I believe your >> on a signed int spoils it somehow.

Heater. · 2013-10-14 02:52

Ah yes, quite so. You would not believe the number of times I forgotten that signed vs unsigned thing over the years when dealing with shifts:)

Of course a >> on a signed number propagates the sign bit down from the top so it can never be the same as a ror. Unless some extra masking is in place to remove propagated bits.
In the rol case we would also be applying a >> to a signed number (the "(a >> (32 - b))" part. That would then require a mask to get rid of the propagated sign bits so it's probably better for the compiler to do what it does.

Heater. · 2013-10-14 03:00

Interesting. Something has changed. With your code I get:

.text
        .balign 4
        .global _rol
_rol
        rol     r0, r1
        lret
        .balign 4
        .global _ror
_ror
        ror     r0, r1
        lret
        .balign 4
        .global _main
_main
        mvi     r0,#1536
        lret

Note the use of "lret". I guess it's the same thing.

Except.. If I compile with "-mcog" then it the same as yours.

Never mind.

Heater. · 2013-10-14 03:05

What we did not say is that it is important to have optimizations switched on to get this rol and ror.
-Os or -O2 or -O3 work.
-O1 does not.

How can I do a circular bit shift in propeller C

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