8051 microcontrollers: 8051 Logical Operations Rotate and Swap Operations Example Programs and Summary

Rotate and Swap Operations

The ability to rotate data is useful for inspecting bits of a byte without using individual bit opcodes. The A register can be rotated one bit position to the left or right with or without including the C flag in the rotation. If the C flag is not included, then the rotation involves the eight bits of the A register. If the C flag is included, then nine bits are involved in the rotation. Including the C flag enables the programmer to construct rotate operations involving any number of bytes.

The SWAP instruction can be thought of as a rotation of nibbles in the A register.

Figure 4.2 diagrams the rotate and swap operations. which are given in the following table:

Mnemonic	Operation
RL A	Rotate the A register one bit position to the left; bit A0 to bit A I, A I to A2. A2 to A3. A3 to A4, A4 to A5, A5 to A6. A6 to A 7, and A 7 to A0
RLC A	Rotate the A register and the carry flag. as a ninth bit, one bit position to the left; bit A0 to bit Al, Al to A2, A2 to A3, A3 to A4, A4 to A5. A5 to A6, A6 to A 7, A 7 to the carry flag, and the carry flag to A0
RR A	Rotate the A register one bit position to the right; bit A0 to bit A 7, A6 to A5, A5 to A4, A4 to A3, A3 to A2. A2 to Al, and At to A0
RRC A	Rotate the A register and the carry flag, as a ninth bit, one bit position to the right; bit A0 to the carry flag. carry flag to A 7, A 7 to A6, A6 to A5, A5 to A4, A4 to A3, A3 to A2, A2 to Al, and Al to A0
SWAP A	Interchange the nibbles of register A; put the high nibble in the low nibble position and the low nibble in the high nibble position

Note that no flags. other than the carry flag in RRC and RLC, are affected. If the carry is used as part of a rotate instruction, the state of the carry flag should be known before the rotate is done.

The programs in this section are written using only opcodes covered to this point in the text. The challenge is to minimize the number of lines of code.

Double the number in register R2, and put the result in registers R3 (high byte) and R4 (low byte).

Thoughts on the Problem The largest number in R2 is FFh; the largest result is 1 FEh. There are at least three ways to solve this problem: Use the MUL instruction (multiply, covered in Chapter 5). add R2 to itself, or shift R2 left one time. The solution that shifts R2 left is as follows:

Note how the carry flag has to be cleared to a known state before being used in a rotate operation.

· Thoughts on the Problem The ports should be input ports for this problem to make any physical sense; otherwise, we would not know whether to use the pin data or the port SFR latch data.

Any time the port is the source of data, the pin levels are read; when the port is the destination, the latch is written. If the port is both source and destination (read-modify-write instructions), then the latch is used.

Find a number that, when X0 Red to the A register, results in the number 3Fh in A.

· Thoughts on the Problem Any number can be in A, so we will work backwards:

Boolean logic, rotate, and swap instructions are covered in this chapter. Byte-level operations involve each individual bit of a source byte operating on the same bit position in the destination byte; the results are put in the destination, while the source is not changed:

Bit-level operations involve individual bits found in one area of internal RAM and certain SFRs that may be addressed both by the assigned direct-byte address and eight individual bit addresses. The following Boolean logical operations may be done on each of these addressable bits:

Write programs that perform the tasks listed using only opcodes that have been discussed in this and previous chapters. Write comments for each line of code and try to use as few lines as possible.

4. Swap the nibbles of R0 and RI so that the low nibble of R0 swaps with the high nibble of RI and the high nibble of R0 swaps with the low nibble of R l .

9. Move bit 4 of RAM location 30h to bit 2 of A. to. XOR a number with whatever is in A so that the result is FFh.

11. Store the most significant nibble of A in both nibbles of register RS: for example, if A = B6h. then RS = BBh.

12. Store the least significant nibble of A in both nibbles of RAM address 3Ch; for example. if A = 36h. then 3Ch = 66h.

13. Set the carry flag to one if the number in A is even; set the carry flag to zero if the number in A is odd.

14. Treat registers R0 and RI as 16-bit registers, and rotate them one place to the left: bit 7 of R0 becomes bit 0 of RI. bit 7 of RI becomes bit 0 of R0, and so on.

18. Shift register B one place to the left: bit 0 becomes a zero. bit 6 becomes bit 7. and so on. Bit 7 is lost.

Mnemonic	Operation
OY A,#OA5h	A = I0l00I0lb = A5h
RR A	A = I I0I00I0b = D2h
RR A	A = 0l I0l00lb = 69h
RR A	A = I0I l0I00b = B4h
RR A	A = 0I0l I0I0b = 5Ah
SWAP A	A= I0I00I0lb = A5h
CLR C	C = 0; A = I0I00I0lb = A5h
RRC A	C = I; A= 0I0I00I0b = 52h
RRC A	C = 0; A = l0I0J00lb = A9h
RL A	A= 0I0I00llb = 53h
RL A	A = I0I00l I0b = A6h
SWAP A	C = 0; A = 0l 101010b = 6Ah
RLC A	C = 0; A = 1 I0I0l00b = D4h
RLC A	C = l; A= I0I0l000b = A8h
SWAP A	C = l; A= l000I0I0b = 8Ah

Mnemonic	Operation
MOY R3,#00h	Clear R3 to receive high byte
CLRC	Clear the carry to receive high bit of 2 x R2
MOY A.R2	Get R2 to A
RLC A	Rotate left, which doubles the number in A
MOY R4,A	Put low byte of result in R4
CLR A	Clear A to receive carry
RLC A
MOY R3.A
The carry bit is now bit 0 of A
Transfer any carry bit to R3

Mnemonic	Operation
MOY A,90h	Copy the pin data from port 1 to A
ORL A,0A0h	OR the contents of A with port 2; results in A
MOY DPTR,#0100h	Set the DPTR to point to external RAM address
MOYX @DPTR.A	Store the result

Mnemonic	Operation
MOV R0,A	Save A in R0
XOR A,#3Fh	XOR A and 3Fh; forming N
XOR A,R0	XOR A and N yielding 3Fh

A= FFh	A XOR 3Fh = C0h	C0h XOR FFh = 3Fh
A= 00h	A XOR 3Fh = 3Fh	3Fh XOR 00h = 3Fh
A= 5Ah	A XOR 3Fh = 65h	65h XOR 5Ah = 3Fh