Shift instructions

Nefastor

Hi all,

I'm using shifts (SHL, SHR) and I am wondering this : suppose I shift left, are the LSB's cleared or set ? (resp, if I shift right, are the MSB's cleared or set ?)

The chip's manual does not indicate this (page 402)

I assume the bits are cleared, since this is the most sensible behavior : SHL can then be used to multiply by powers of 2.

Still, I'd prefer to be sure

Jean

asterick

If it's a logical shift, than the MSB's are copied from the previous MSB.

otherwise, 0's are shifted in.

Nefastor

asterick said...
If it's a logical shift, than the MSB's are copied from the previous MSB.

otherwise, 0's are shifted in.

Not sure you got this right : a shift where the MSB is copied from the previous MSB would be a right arithmetic shift, not a right logical shift : keeping the MSB preserves the sign bit.

My question still stands.

Jean

rokicki

LSBs are cleared on left shifts. On logical shift rights, MSBs are cleared. On arithmetic shift rights, MSBs are copied.

Paul Baker

All shift routines will shift in a 0, except arithmetic shift right which shifts the MSB in.

Mathmatically its:
shl = multiply by two
shr = divide by two unsigned
sar = divide by two signed

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

Post Edited (Paul Baker (Parallax)) : 1/3/2007 10:19:36 PM GMT

asterick

Yea, I had it backwards.

rokicki

The tricky case is the extended rotates. These copy the carry bit to *all* "inshifted" positions, rather than
doing an extended (33-bit) "rotate".