The bit shifting operations are fairly straightforward.<\/p>\n
; arithmetic (signed) shifts\r\n SHAL Rn ; Rn <<= 1, T = the bit shifted out\r\n SHAR Rn ; Rn >>= 1, T = the bit shifted out\r\n\r\n ; logical (unsigned) shifts\r\n SHLL Rn ; Rn <<= 1, T = the bit shifted out\r\n SHLR Rn ; Rn >>= 1, T = the bit shifted out\r\n SHLL2 Rn ; Rn <<= 2\r\n SHLR2 Rn ; Rn >>= 2\r\n SHLL8 Rn ; Rn <<= 8\r\n SHLR8 Rn ; Rn >>= 8\r\n SHLL16 Rn ; Rn <<= 16\r\n SHLR16 Rn ; Rn >>= 16\r\n<\/pre>\nYou 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<\/code> and aSHLL<\/code>.<\/p>\nNote also that
SHAL<\/code> andSHLL<\/code> are functionally equivalent. But they have different encodings, so the designers burned an opcode for a redundant operation.<\/p>\nThere are no “large shift” options for right shifts. You can perform multiple one-bit shifts, or use a variable shift:<\/p>\n
SHAD Rm, Rn ; if Rm > 0: Rn <<= (31 & Rm)\r\n ; if Rm = 0: nop\r\n ; if Rm < 0: Rn >>= (31 & -Rm), signed\r\n\r\n SHLD Rm, Rn ; if Rm > 0: Rn <<= (31 & Rm)\r\n ; if Rm = 0: nop\r\n ; if Rm < 0: Rn >>= (31 & -Rm), unsigned\r\n<\/pre>\nNote that these shift instructions shift both left and<\/i> 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.<\/p>\n
Finally, we have rotation.<\/p>\n
ROTL Rn ; rotate left, T contains carried-out bit\r\n ROTR Rn ; rotate right, T contains carried-out bit\r\n ROTCL Rn ; 33-bit rotate through T\r\n ROTCR Rn ; 33-bit rotate through T\r\n<\/pre>\nThe 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<\/var>. For the 33-bit rotations, the T<\/var> flag acts as the 33rd bit.<\/p>\n
We saw earlier that there is no
NEGV<\/code> instruction. To detect overflow from a negation, you just have to check for the value0x80000000<\/code> directly. Here’s the shortest sequence I could come up with:<\/p>\n; branch if Rn equals 0x80000000\r\n rotl Rn ; rotate left one bit\r\n dt Rn ; decrement and test for zero\r\n bt underflow ; Y: underflow occurred\r\n<\/pre>\nThe result of the
DT<\/code> is zero if the previous value was 1, and the previous value was 1 if the original value was0x80000000<\/code>.<\/p>\n