Pl INSTRUC2.DOC
{Languages|Pl INSRUC2.DOC}}
CONTINUATION OF THE COURSE ON INSTRUCTIONS from the INSTRUC.DOC file
LSL #BBB,dn (.B),[.W],(.L) (Logical Shift Left) -------------- or LSL dm,dn (.B),[.W],(.L) -------------- or LSL.W destination (.W) ------------------ Performs a shift of #BBB or dm (depending on the syntax) BITS of the data register dn to the left. The replacement bits are ZEROS. It is necessary that 1<= #BBB <= 8 for the 1st form. ------------- For the 3rd form, the allowed addressing modes for the destination operand are: ----------- BBBB BB (an) -(an) (an)+ d(an) d(an,rn) But in this case, the operation can only shift 1 SINGLE BIT. For example, if d0=%10000101110001101111011100000110 (word) ------ And if I write LSL.W #1,dn or MOVE.W #1,d1 LSL.W d1,d0, we get: d0=%00001011100011011110111000001100 <-------- <--:0 All the bits of the lower WORD of the data register dn (.W) have been shifted by 1 bit to the left: The least significant bit entering is always replaced by a NULL bit. Meaning: #BBB or dm BITS are shifted out of dn to the left and the replacement bits entering on the right of dn are NULL BITS. The contents of dn are therefore changed, the MSB is also likely to change, that's why the SIGN of dn may not necessarily be preserved. The CCR will change according to the new value of dn. --- N=1 if the MSB is 1, otherwise it is 0 Z=1 if all bits of dn are null, otherwise it is 0 V is always set to 0 C and X are set to the value of the LAST BIT EXITED from dn. Example of use: ---------------------- MOVE #%1111111101010101,d0 LSL.B #5,d0 A shift of 5 bits of the lower BYTE of d0 is operated (only bits 0 to 7 of d0 are therefore concerned by lsl.B) We get: CCR --------<------<:0 X=C=0 <- 1111111110100000 in d0 5 null bits have entered the right of the lower byte of d0, the last bit to exit from d0 was null: bit X=C=0 in the CCR LSR #BBB,dn (.B),[.W],(.L) (Logical Shift Right) -------------- or LSR dm,dn (.B),[.W],(.L) -------------- or LSR.W destination (.W) ------------------ Performs a shift of #BBB or dm (depending on the syntax) BITS of the data register dn to the right. The replacement bits are ZEROS. It is necessary that 1<= #BBB <= 8 for the 1st form. ------------- For the 3rd form, the allowed addressing modes for the destination operand are: ----------- BBBB BB (an) -(an) (an)+ d(an) d(an,rn) But in this case, the operation can only shift 1 SINGLE BIT. For example, if d0=%10000101110001101111011100000110 (word) ------ And if I write LSR.W #1,dn or MOVE.W #1,d1 LSR.W d1,d0, we get: d0=%01000010111000110111101110000011 0:--> -------> All the bits of the lower WORD of the data register dn (.W) have been shifted by 1 bit to the right: The most significant bit entering is always replaced by a NULL bit. Meaning: #BBB or dm BITS are shifted out of dn to the right and the replacement bits entering on the left of dn are NULL BITS. The contents of dn are therefore changed, the MSB is also likely to change, that's why the SIGN of dn may not necessarily be preserved. The CCR will change according to the new value of dn. --- N=1 if the MSB is 1, otherwise it is 0 Z=1 if all bits of dn are null, otherwise it is 0 V is always set to 0 C and X are set to the value of the LAST BIT EXITED from dn. Example of use: ---------------------- MOVE #%1111111101010101,d0 LSR.B #5,d0 A shift of 5 bits of the lower BYTE of d0 is operated. (only bits 0 to 7 of d0 are therefore concerned by lsr.B) We get: ------0:>------> CCR d0 :1111111100000010 -> X=C=1 5 null bits have entered the left of the lower byte of d0, the last bit to exit from d0 was 1: bit X=C=1 in the CCR Back to ASM_Tutorial