Bit shifting left on P1 PASM

forthnutter

My goal was get the status bit 30 in a variable.
I used SHL instruction so it goes into C flag.

if W has a value 0x80000000

SHL W,#2 WC

The C flag would always equal 1
The Propeller Manual says the C flag is set equal to Value’s original bit 31.
Does that mean shift of the number of bits, except 0, would result in C equal to 1 because bit 31 was set at the start of instruction ?.
My mind was in 68000 mode, thinking the number of bits shifted would go into C flag would work, but no.

A work around was to do two shl
SHL W,#1
SHL W,#1 WC

Wuerfel_21

The C flag is always set to the original MSB of the shifted register, even if your shift count is zero. (same applies to right shifts, except they give you the LSB)

JonnyMac

If you have other bits to check, you could create a mask (at the end of PASM code, before RES declarations)

BIT30                   long      1 << 30

then use AND to check it. With only one bit in the mask, C will contain the state of the tested bit.

                        and       t1, BIT30             wc, nr

With NR, the tested variable is not modified.

forthnutter

@JonnyMac said:
If you have other bits to check, you could create a mask (at the end of PASM code, before RES declarations)

BIT30                   long      1 << 30

then use AND to check it. With only one bit in the mask, C will contain the state of the tested bit.

                        and       t1, BIT30             wc, nr

With NR, the tested variable is not modified.

Thanks for that, I did not think of doing that way.
I suppose the TEST instruction does the same thing.

Wuerfel_21

Yes, TEST is identical to AND with NR flag. (conversely, you can write TEST with a WR flag to get AND). I find TEST easier to read.

JonnyMac

I suppose the TEST instruction does the same thing.

Indeed -- and I should have remembered that given I used the word 'test' in my reply!

I find TEST easier to read.

Agree.

                        test      t1, BIT30             wc