Download The Assembly Language of the Box-5 of Computer Architecture | CPSC 5155G and more Study notes Computer Architecture and Organization in PDF only on Docsity!
The Assembly Language of the Boz–
The Boz–5 uses bits 31 – 27 of the IR as a five–bit opcode.
Of the possible 32 opcodes, only 26 are implemented. Op-Code Mnemonic Description 00000 HLT Halt the Computer 00001 LDI Load Register from Immediate Operand 00010 ANDI Logical AND Register with Immediate Operand 00011 ADDI Add Signed Immediate Operand to Register 00100 NOP Not Yet Defined – At Present it is does nothing 00101 NOP Not Yet Defined – At Present it is does nothing 00110 NOP Not Yet Defined – At Present it is does nothing 00111 NOP Not Yet Defined – At Present it is does nothing
This strange gap in the opcodes caused by not using 4, 5, 6, and 7 is an example
of adjusting the ISA to facilitate a simpler design of the control unit.
With this grouping, the instructions with a “00” prefix either have no argument
or have an immediate argument; in short, no memory reference.
More Opcodes
Here are the opcodes in the range 8 to 15.
Op-Code Mnemonic Description 01000 GET Input from I/O Device Register to Register 01001 PUT Output from Register to I/O Device Register 01010 RET Return from Subroutine 01011 RTI Return from Interrupt (Not Implemented) 01100 LDR Load Register from Memory 01101 STR Store Register into Memory 01110 JSR Call a subroutine. 01111 BR Branch on Condition Code to Address
Here we begin to see some structure in the ISA.
The “01” prefix is shared by all instructions that generate memory address
references. This grouping simplifies the design of the Major State Register,
which is an integral part of the control unit.
Syntax of the Assembly Language
Immediate Operations
The syntax of the immediate operations is quite simple.
Syntax Example
LDI %RD, value LDI %R3, 100
ANDI %RD, %RS, value ANDI %R5, %R3 0xFFA
ADDI %RD, %RS, value ADDI %R5, %R2, 200
NOP NOP
Most implementations of immediate addressing have only one register in the
machine language instruction. Our implementation with two registers is again
due to considerations of simplifying the control unit.
In the AND Immediate instruction, the immediate operand is zero extended to
32 bits before being applied. Let R3 contains the 32–bit number 0x77777777.
R3 0111 0111 0111 0111 0111 0111 0111 0111
FFA08 0000 0000 0000 1111 1111 1010 0000 1000
Result 0000 0000 0000 0111 0111 0010 0000 0000
Syntax of the Assembly Language (Part 2)
Input / Output Operations
Syntax Example
GET %Rn, I/O_Reg GET %R2, XX
// Load the general-purpose register from
// the I/O device register.
PUT %Rn, I/O_Reg PUT %R0, YY
// Store the contents of register into the
// I/O device register.
Load/Store Operations
The syntax of these is fairly simple. For direct addressing we have the
following.
Syntax Example
LDR %Rn, address LDR %R3, X Loads %R3 from address X
STR %Rn, address STR %R0, Z Loads %R0 into address Z
This clears M[Z]
Syntax of the Assembly Language (Part 4)
Unary Register-To-Register 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 – 0
Op Code Destination
Register
Source
Register
5–bit Shift Count
Not
Used
Opcode = 10000 LLS Logical Left Shift
10001 LCS Circular Left Shift
10010 RLS Logical Right Shift
10011 RAS Arithmetic Right Shift
10100 NOT Logical NOT (Shift count ignored)
NOTES: 1. If (Count Field) = 0, a shift operation becomes a register move.
2. If (Source Register = 0), the operation becomes a clear.
3. Circular right shifts are not supported, because they may be
implemented using circular left shifts.
4. The shift count, being a 5 bit number, has values 0 to 31 inclusive.
Shift Examples
Here are figures showing the varieties of right shifts on signed 8–bit integers.
Syntactic Sugar
Syntactic Sugar Assembled As Comments CLR %R2 LDI %R2, 0 Clears the register CLW X STR %R0, X Clears the word at address X INC %R2 ADDI %R2, 1 Increments the register DEC %R2 ADDI %R2, -1 Decrements the register NOP LLS %R0, %R0, 0 No-Operation: Does Nothing RCS %R3, %R1, 3 LCS %R3, %R1, 29
Right circular shift by N is the same as left circular by (32 – N). The shift count
must be a constant number.
DBL %R2 LLS %R2, %R2, 1 Left shift by one is the same as a multiply by 2. MOV %R2, %R3 LSH %R2, %R3, 0 Shift by 0 is a copy. NEG %R4, %R5 SUB %R4, %R0, %R5 Subtract from %R0 0 is the same as negation.
Syntactic Sugar (Part 2)
Syntactic Sugar Assembled As Comments TST %R1 SUB %R0, %R1, %R
This compares %R1 to zero by subtracting %R0 from it, discarding the result.
CMP %R1, %R2 SUB %R0, %R1, %R
Determines the sign of(%R1) – (%R2), discarding the result.
BRU BR 000 Branch always BLT BR 001 Branch if negative BEQ BR 010 Branch if zero BLE BR 011 Branch if not positive BCO BR 100 Branch if carry out is 0 BGE BR 101 Branch if not negative BNE BR 110 Branch if not zero BGT BR 111 Branch if positive
