CPU Architectures Motorola 68000 M68000 Data

CPU Architectures Motorola 68000

• Several CPU architectures exist currently: • Arbitrary pick: Motorola 68000 – Motorola (or m68k) – Intel • CISC processor – AMD (Advanced Micro Devices) – translates instructions – PowerPC into microcode, and • Pick one to study; others will be variations on this. executes a sequence of micro-instructions on a RISC architecture. • 32-bit architecture

– all data is 32 bits long. 4 – addresses are 24-bits, Select stored in 32-bit values. • Datapath shown in diagram.

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M68000 Data

• Data is stored in “big endian ” format – most significant byte is stored in least significant segment of address. “Little endian ” stores the least significant byte in the least significant portion of specified address. – Mnemonic: “big end in”. – Example: Data value 4A3B2C1D at memory location 0x1000 • Big endian: 1000 1001 1002 1003 4A 3B 2C 1D

• Little endian: 1000 1001 1002 1003 1D 2C 3B 4A • A word in M68000 is 16 bits (2 bytes), and a long word is 32 bits (4 bytes). – Most data and address accesses are long word operations.

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1 Other M68000 Updates How It All Works

• MAR and MDR pretty much the same. • Basic operation of the M68000 involves performing – uses signal called MFC (Memory Function Completed ) which one or more of the following functions: is sent to the CPU after a memory operation is done. 1. Fetching contents of a memory location and storing these • 8 data registers (R0 to R7) and 8 address registers contents in a CPU register. (A0 to A7) 2. Storing the contents of a CPU register into a location in – A7 is also known as SP , since it stores the stack pointer . memory. • Access to the components of the CPU (i.e. the arrows 3. Transferring the contents of one CPU register to another. in the diagram) is controlled by a latched input . 4. Performing an arithmetic or logical operation and storing – The instruction decoder controls which latches are activated. the result in a CPU register. – The operation performed by the ALU is also controlled by the instruction decoder. – The result of the ALU’s operations (e.g. C, V, N) are • Let’s look at how to do some of these things… conveyed back to the instruction decoder through a status register, usually located in the decoder itself.

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Basic Control Signals Basic Control Signals Internal processor bus Internal processor • “Add the contents of R0 and R1, bus • To perform add operation, Riin and store the result in R0.” Riin perform the following steps: Ri • Instruction form: Add R1, R0 Ri 1. get R1 (or R0) into Y Riout

Riout 2. get R0 (or R1) onto bus Yin Yin 3. tell ALU to perform addition • Need to activate addition 4. store result in Z Y mode in the ALU, and get Y Constant 4 Constant 4 5. put contents of Z onto bus the contents of R0 and R1 6. store contents of bus in R0 Select MUX Select MUX into inputs A & B at the • Of these steps, which have to be A B A B ALU same time. ALU done simultaneously in order for – Problem: Single-bus architecture Zin Zin the operation to succeed? only allows one value on the bus. Z Z – number of grouped steps determines – Solution: Use the Y register to store the number of clock cycles per Z out Z out a value temporarily. instruction.

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2 Basic Control Signals Basic Control Signals Internal processor bus • “Fetch the next instruction” • Translate grouped steps into control Riin – Required part of every instruction. signals from instruction decoder: Ri – Two sets of steps needed… – Step 1: R1 out , Y in Riout • Operation #1: Fetch instruction – Step 2: R0 out , Select, Add, Zin Yin 1. Put PC value into MAR – Step 3: Zout , R0 in Y 2. Fetch instruction from memory • Note: The operation “Add R1, R0” Constant 4 3. Store retrieved instruction in IR is like pseudocode for CPUs. Select MUX • Operation #2: Increment PC We will use this for the next little A B while, but assembly language is ALU 1. Increment PC by 4 4 typically used for implementing Zin 2. Store incremented value back in PC Select these instructions in real-life Z • Can these two operations be

programs Zout performed in parallel? – e.g. ADD.W D1, D0

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Basic Control Signals Instruction Retrieval FAQ

• Translate operations into control signals: • What is the “Read” signal, exactly? – Fetch instruction – Sends signal to memory, requesting the data value stored at the MAR’s address. When complete, the MDR will be storing 1. PC , MAR , Read, WMFC PC (original) out in the retrieved data. 2. MDR , IR IR (updated) out in – “Write” is the only other memory operation, which stores the – Increment PC MDR’s value into the memory location specified by the MAR. 1. PC , Add, Z out in PC (original) • Why does WMFC get moved to the second step? 2. Z , PC PC (updated) out in – WMFC suspends CPU activity until the memory access is • To determine how these could be done in parallel, complete. Typically, this is much larger than any single CPU examine what signals are on the bus in each step… instruction, so get everything else done before you WMFC. • How does the PC get incremented? – The mux either adds the Y value to the bus (if Select is high) 1. PC out , MAR in , Read, Add, Zin 2. Z , PC , WMFC or just adds 4 (if Select is low). Instructions are 32 bits long, out in and each memory location stores 1 byte. You do the math. 3. MDR out , IR in

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3 Instruction Retrieval Shortcuts Impact on Timing • The structure of the datapath has a direct influence • Since this instruction retrieval step is so common, on the clock speed of the CPU. designers try to find ways to shorten the steps. • Consider the delays involved in the Add operation: – Memory access is always horribly slow, but that can’t be – Activation time for tri-state buffer on R0 out helped. Advances in semiconductor devices and memory – Transmission time for R0 onto bus technology are needed for this (see Moore’s Law). out – Propagation delay through ALU – PC increments offer potential design enhancements: – Setup time for loading value into Z register • CPU designs sometimes create a dedicated increment circuit for the PC, so that it doesn’t use the bus or involve the ALU. – Hold time for Z before lowering R0 out .

• If the PC stores the instruction address divided by 4, the PC can Turn-on Preparation time for Transmission Propagation Setup Hold increment by one instead of 4 for each new instruction. The PC delay time for next 3-state time time time instruction will then be multiplied by 4 before being copied to the MAR. driver through ALU • The PC and IR can have a direct connection to memory, much like the MAR and MDR. R0 out • Instruction memory can be made separate from main memory. Zin

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Addressing Modes Addressing Modes

• While we’re talking about the Add operation, how – Autoincrement mode = a version of register indirect mode, many ways are there to specify operands? where the address stored in the register is incremented by 1, 2 or 4 after the memory access, depending on the length – Register direct mode = what we’ve seen so far. of the data being accessed.

Add R6, R0 Add (A0)+, R0

– Register indirect mode = the register stores the memory – Autodecrement mode = same as autoincrement, only the address of the operand. address is decremented before the memory access.

Add (A0), R0 Add -(A0), R0

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4 Addressing Modes Addressing Modes

– Immediate mode = the value of the operand is given in the – Basic index mode = a 16-bit signed offset and an address instruction itself. The ‘#’ character denotes an immediate register are specified in the instruction. The sum of the offset value. and the register’s contents specifies the operand location.

Add #200, R0 Add 42(A0), R0

– Absolute mode = the absolute address to be used in the – Full index mode = an 8-bit signed offset, an address register operation is found in the instruction itself. The ‘$’ character and a data register are specified in the instruction. The sum denotes that the address is in hexadecimal notation (0x…). of the offset and the two registers specifies the operand • Absolute value can be either short or long . If long, then 24 bits location. are given. If short, 16 bits are given and then sign extended. Add 53(A0, R1), R0 Jmp $1200

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Addressing Modes Instruction Types

– Basic relative mode = same as basic index mode, only using • Now that we’ve specified the different forms that the program counter instead of an address register. operands can take, what about the operations? • Operations for the M68000 fall into one of three Add 42(PC), R0 major categories: – Data manipulation (arithmetic, shifting, copying) – Control flow (branching, jumping, subroutines) – Full relative mode = same as full index mode, only using the program counter instead of an address register.

Add 53(PC, R1), R0

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5 Data Manipulation Instructions Data Manipulation Instructions – Shifting • 5 principal methods for manipulating data. LSL (logical shift left) LSR (logical shift right) – Arithmetic : ASL (arithmetic shift left) ASR (arithmetic shift right) ROL (rotate left) ROR (rotate right) ADD (addition) SUB (subtraction) ROXL (rotate left through X) ROXR (rotate right through X) MULU (unsigned multiplication) MULS (signed multiplication) DIVU (unsigned division) DIVS (signed division) – Copying NEG (arithmetic negation) CMP (comparison) MOVE (move value) MOVEA (move address) EXG (exchange values) LEA (load address) – Logical : AND (logical AND) OR (logical OR) – Binary Arithmetic NOT (complement) EOR (exclusive OR) ABCD (add BCD values) SBCD (subtract BCD values)

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Control Flow Instructions Control Flow Instructions • Branch condition codes

• 3 main types of control flow instructions. Machine Code Condition Suffix (cc) Name Test Condition 0000 T True Always true 0001 F False Always false – Branching: 0010 HI High C ∨ Z = 0 0011 LS Low or Same C ∨ Z = 1 BRA (branch always) Bcc (branch conditionally) 0100 CC Carry Clear C = 0 BSR (branch to subroutine) DBcc (decrement, and Bcc ) 0101 CS Carry Set C = 1 0110 NE Not Equal Z = 0 0111 EQ Equal Z = 1 – Jumping: 1000 VC oVerflow Clear V = 0 JMP (jump to address) JSR (jump to subroutine) 1001 VS oVerflow Set V = 1 1010 PL Plus N = 0 1011 MI Minus N = 1 – Subroutines: 1100 GE Greater or Equal N ⊕ V = 0 1101 LT Less Than N ⊕ V = 1 RTS (return from subroutine) RTE (return from exception) 1110 GT Greater Than Z ∨ (N ⊕ V) = 0 1111 LE Less or Equal Z ∨ (N ⊕ V) = 1

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6 Data Details Other Important Commands

• When moving or modifying data, it’s important to • There are many more commands than the ones listed specify the length of the data being moved or here, but most are rather obscure and rarely used. modified. • A few important ones commands that were missed: • Example: move.b (a0), r1 – dc (declare constant) = allocated space for a constant value, – This moves the byte at a0 ’s location into the least significant which can be referred to in place of a data value. r1 byte of . • Example: value: dc.b 42 – It’s important to consider the endian of the system you’re move.w value, r0 using (see slide 4). • In this case, value is like a global variable to the program. • Also, register names can vary from system to system. – org (program origin) = set a location in memory for the – M68000 often uses d0-d7 to represent data registers, e.g. following assembly instructions should be written. – Don’t get confused. • Example: org $1000 • This starts the following code at location 1000 (hex) in memory.

Other Important Commands Other Trap Details

– trap (software interrupt) = call an interrupt subroutine, • The first 256 long words in memory store the such as printing to the screen, or receiving data from memory addresses of vital processor subroutines. standard input. – address 0 contains the address of the initial startup routine. • Example: trap #14 • This is also called the exception vector table . • This calls the subroutine that goes to trap vector #14, which loads whatever address is at memory location 14, and jumps • Locations 32-47 of this table are allocated to the trap the program counter to that address (usually in read-only vectors , and the contents of these entries can be set memory). • Often, the program that resides at the address in vector #14 is by the operating system (which you create). responsible for printing characters to the screen (m68k only). • Calling Trap #14 would then look • Before jumping, the current PC and SP values are pushed onto up entry 46 in the exception table, the stack. The RTE instruction pops these value back. go to the address specified in that • Other trap locations will be responsible for other operations (e.g. handling misaligned word addresses, reading from entry, and execute the subroutine keyboard, etc.) at that address.

7 Memory-Mapped I/O Memory-Mapped I/O

• The trap entries in the exception table contain the • When performing read or write operations, we’re typically interested in the parallel or serial ports . address of the exception code. How does that – Printer/monitor parallel exception code actually read or write from external – Keyboard/internet serial devices? • To perform these actions, the M68xxx systems either • The trap code assumes that the information to write is map specific devices to location in memory, or will in a particular register, or that the information to read implement a multi-function peripheral (MFP) interface . should be stored in a particular data register. • The MFP interface is a series of registers in memory – In the case of reading or writing strings, an address register that are responsible for the various aspects of I/O is used instead to store the initial address of the string. A null operations: character is used to indicate that the string has terminated. – Data direction • The actual address of the device’s IO location can – Control register vary, depending on the hardware implementation. – Receiver/transmitter status register – Data register

Multi-Function Peripherals Opcodes

• Each register in the MFP interface is assumed to be an • When an assembly language program is compiled by odd offset from a given location in memory. the assembler, it is translated into binary instructions – Let’s assume that it’s at $E70000 (various sources give that can be recorded in memory and understood by different values for this starting address) the instruction decoder. • General-Purpose I/O Register (GPIP) offset of 1 – Also called opcodes (short for operation codes ) • Universal Synchronous/Asynchronous Receiver/Transmitter (USART) Control Register (USC) offset of 0x29 • How does it do this translation? What do these binary • Receiver Status Register (RSR) offset of 0x2B instructions look like? • Transmitter Status Register (TSR) offset of 0x2D • Each instruction can be broken down into parts, • USART Data Register (USR) offset of 0x2F – to see what assembly language instruction it corresponds to – To perform an I/O operation, use the control registers to set – to see how the instruction decoder understands its contents up the nature of the operation (synchronous, byte/word/long), and then use the data and status registers to send information and verify it was received.

8 Opcodes Opcodes

• Note: See Appedix C • The instruction decoder reads Bits 15-12 Operation in the text and the the first 4 bits, and knows 0000 Bit manipulation/MOVEP/Immed 0001 Move Byte M68000 reference more about how to handle 0010 Move Long manual on the web the instruction from there. 0011 Move Word 0100 Miscellaneous – Example: 0010 Move Long page for opcode tables and more opcode information. 0101 ADDQ/SUBQ/Scc/DBcc/TRAP • The opcode is broken down into sections differently, • Sometimes, a fifth bit is 0110 Bcc/BSR/BRA depending on the type of instruction. needed to distinguish 0111 MOVEQ 1000 OR/DIV/SBCD • In order to know how to break down the instruction, instructions within the 1001 SUB/SUBX the first section of the instruction contains the general categories. 1010 (Unassigned, Reserved) 1011 CMP/EOR – Example: 1011 identifier for the instruction type. 1100 AND/MUL/ABCD/EXG – most significant 4 bits (15-12) of a 16-bit instruction. • if bit 8 is 0 CMP 1101 ADD/ADDX • if bit 8 is 1 EOR 1110 Shift/Rotate/Bit Field 1111 MC68040 & CPU32 Extensions

Opcodes Opcodes

• After the instruction has been identified, the • Consider the first opcode format for this instruction: remaining 16 bits can be interpreted according to the tables in Appendix C from the text. 1101 DDD1 SS EEEEEE • Example: ADD • This instruction corresponds to: ADD Dn, X – Dn is a data register (n is a value from 0 to 7) –X is the other operand, which also acts as the destination • What do the letter fields mean? – DDD = Data register number (i.e. Dn ) • This table entry specifies the operand types and sizes – SS = Size ( 00 byte, 01 word, 10 long word) for this instruction, the format for the opcode, the – EEEEEE = Effective address of destination operand (i.e. X) underlying semantics of the instruction, and the • First three bits of EEEEEE addressing mode status register codes after the instruction takes place. • Second three bits of EEEEEE register number (if applicable)

9 Opcodes Special Opcode Cases

• So given the format Addressing Mode Mode Field Register Field 1101 DDD1 SS EEEEEE Data register direct 000 Register Number • Branching instructions how would you Address register direct 001 Register Number – Branch instructions require an offset to tell them how far specify the instruction Address register indirect 010 Register Number forward or backward to branch. Autoincrement 011 Register Number ADD.W R1, (A6) – Keep in mind that the PC will have incremented already Autodecrement 100 Register Number (slides 10 & 11), so the address in the program counter will in opcode format? Indexed basic 101 Register Number be one instruction larger than expected. Indexed full 110 Register Number • Specify parts: • Supplemental information – DDD = 001 Absolute short 111 000 Absolute long 111 001 – Some instructions (such as JMP) will involve an address or – SS = 01 Relative basic 111 010 other information that is too big to fit into the instruction – EEEEEE : Relative full 111 011 itself.

• Address mode = 010 Immediate or SR 111 100 – When this happens, the instruction decoder understands • Register no. = 110 that the necessary information can be found in the 16 or 32 • Result? bits following the previous instruction, and will load that 1101 0011 01 010110 information into a data register instead of the IR.