The SuperH-3, part 8: Bit shifting

The bit shifting operations are fairly straightforward.

    ; arithmetic (signed) shifts
    SHAL Rn         ; Rn <<= 1, T = the bit shifted out
    SHAR Rn         ; Rn >>= 1, T = the bit shifted out

    ; logical (unsigned) shifts
    SHLL Rn         ; Rn <<= 1, T = the bit shifted out
    SHLR Rn         ; Rn >>= 1, T = the bit shifted out
    SHLL2 Rn        ; Rn <<= 2
    SHLR2 Rn        ; Rn >>= 2
    SHLL8 Rn        ; Rn <<= 8
    SHLR8 Rn        ; Rn >>= 8
    SHLL16 Rn       ; Rn <<= 16
    SHLR16 Rn       ; Rn >>= 16

You cannot shift by arbitrary constant amounts. Only certain fixed values are permitted. If you want to shift left by, say, 9, you’ll have to construct it from a SHLL8 and a SHLL.

Note also that SHAL and SHLL are functionally equivalent. But they have different encodings, so the designers burned an opcode for a redundant operation.

There are no “large shift” options for right shifts. You can perform multiple one-bit shifts, or use a variable shift:

    SHAD Rm, Rn     ; if Rm > 0: Rn <<= (31 & Rm)
                    ; if Rm = 0: nop
                    ; if Rm < 0: Rn >>= (31 & -Rm), signed

    SHLD Rm, Rn     ; if Rm > 0: Rn <<= (31 & Rm)
                    ; if Rm = 0: nop
                    ; if Rm < 0: Rn >>= (31 & -Rm), unsigned

Note that these shift instructions shift both left and right, depending on the sign of the shift amount. If you want to shift right by an amount in a register, you therefore need to negate the value, and then shift left.

Finally, we have rotation.

    ROTL Rn        ; rotate left, T contains carried-out bit
    ROTR Rn        ; rotate right, T contains carried-out bit
    ROTCL Rn       ; 33-bit rotate through T
    ROTCR Rn       ; 33-bit rotate through T

The rotation instructions rotate either a 32-bit or 33-bit value by one position. For the 32-bit rotations, the bit that rotated off the end is copied to T. For the 33-bit rotations, the T flag acts as the 33rd bit.

We saw earlier that there is no NEGV instruction. To detect overflow from a negation, you just have to check for the value 0x80000000 directly. Here’s the shortest sequence I could come up with:

    ; branch if Rn equals 0x80000000
    rotl Rn        ; rotate left one bit
    dt   Rn        ; decrement and test for zero
    bt   underflow ; Y: underflow occurred

The result of the DT is zero if the previous value was 1, and the previous value was 1 if the original value was 0x80000000.

This is a destructive operation, so do it in a scratch register. You should have one available, since it’s the source register for the NEGV you were checking.

We’ll look more at constants next time.