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CSE243: Introduction to Computer Architecture and Hardware/Software Interface Topics covered: 68000 Instruction Set Architecture 68000 Instruction Set Architecture Registers and addressing Addressing modes Instructions Assembly language Branch instructions Stacks and subroutines Simple programs in 68000 assembly language 1 Registers and addressing External word length: Word length used to transfer data between the processor and memory. 16 bits. Internal word length: Refers to the size of the processor registers 32 bits. 2 Register structure 8 data registers (D0-D7) and 8 address registers (A0-A7): Each is 32 bits long. Instructions use operands of three different lengths: Long word – 32 bits Word – 16 bits Byte – 8 bits A byte or a word operand stored in a data register is in the low order bit positions of a register: Most instructions operating on byte or word operands do not affect the higher order bit positions. Some instructions extend the sign of a shorter operand into the high-order bits. 3 Register structure (contd..) Address registers hold information used in determining the addresses of memory operands: Address and data registers can also be used as index registers. Address register A7 has a special function of serving as the processor stack pointer. Address registers and calculations involve 32 bits: Only the least 24 significant bits of an address are used externally to access the memory. Processor Status Register (SR): Five condition code bits. Three interrupt bits (function will be clear when we discuss I/0) Two mode-select bits. 4 Register structure (contd..) Long word Word Byte 31 16 15 8 7 0 D0 D1 D2 D3 D4 User stack pointer A7 Stack pointers Supervisor stack pointer Data registers PC D5 15 13 10 8 D6 SR T - Trace mode select S - Supervisor mode select I - Interrupt mask D7 A0 4 0 Status register CVZN- Carry Overflow Zero Negative A1 A2 A3 Address registers A4 A5 A6 5 Addressing Word addresses Contents 0 byte 0 byte 1 2 byte 2 byte 3 i byte i byte i+1 i+2 byte i+2 byte i+3 224 - 2 byte 224 - 2 byte 224 - 1 Long word 0 Long word i •Byte addressable memory •Memory is organized in 16-bit words. •Two consecutive 16-bit words constitute one 32-bit long word. •Word address must be an even number, that is, words must be aligned on an even boundary. •Byte in the high-order position has the same address as the word, the byte in the low-order position has the next higher address. This is the big endian address assignment. 6 Addressing (contd..) 68000 generates 24-bit memory addresses. Addressable space is 224 (16,777,216 or 16M) bytes. Addressable space may be thought of as consisting of: 512 (29) pages of 32K (215) bytes each. Page 0 is hexadecimal addresses 0 to 7FFF. Page 1 is hexadecimal addresses 8000 to FFFF. Last page is hexadecimal addresses FF8000 to FFFFF. 7 Addressing modes Instruction size: Many instructions are one word (16-bit long), some require additional 16-bit words for extra addressing information. First word of an instruction is the OP-code word. Subsequent words (if any) are called extension words. Note that the 68000 does not place a one word limit on the instruction size, it is a CISC architecture. Immediate mode Operand is included in the instruction. Byte, word, long-word operands are specified in the extension word. Some very small operands are included in the OP-code word. 8 Addressing modes (contd..) Absolute mode: Absolute address of an operand is given in the extension word. Long mode: 24-bit address is specified. Short mode: 16-bit value is specified, which serves as the loworder 16 bits of an address. • Sign bit of the 16-bit value is extended to provide the highorder 8 bits. • Sign-bit is 0 or 1, so two pages can be addressed: 0, and FF8. Register mode: Operand is in a processor register specified in the instruction. Register indirect mode: Effective address (EA) of an operand is in a processor register which is specified in an instruction. 9 Addressing modes (contd..) Autoincrement mode: EA is in one of the address registers which is specified in the instruction. After the operand is addressed, the contents of the address register are incremented by 1 (byte), 2 (word), 4 (long-word). Autodecrement mode: Same as above, except the contents are decremented before the operand is accessed. Basic index mode: 16-bit signed offset and an address register are specified. Full index mode: 8-bit signed offset, an address register and an index register (either a data or address register) are specified. Basic relative and Full relative modes: Same as Basic index and Full index modes respectively, except PC is used instead of address register. 10 Addressing modes (contd..) 1000 OP-code word 1002 Extension word 100 = offset 1102 1104 6 = index 1106 1108 Operand 6 A1 •Consider the instruction: ADD 100(PC,A1), D0 •Full relative mode for the source operand. •Register mode for the destination operand. •Compute the EA of the source operand: - Suppose A1 holds the value 6. - Suppose the OP-code word is at 1000. Array - After the OP-code word is fetched, PC points to the extension word, i.e. 1002. EA = [PC] + [A1] + 100 Thus, EA is 1108. •This could be used to address the data in an array form. •In relative modes, PC need not be specified explicitly in the instruction. 11 Instructions Extensive set of instructions. Both two-operand and one-operand instructions. Two-operand instructions: Format: OP src, dst. Assembly language notation, the actual encoding may not be in the same order. OP specifies the operation performed. Result is placed in the destination location. For most two operand instructions, either src, or dst operand must be in a data register. Other operand may be specified using any one of the addressing modes. The only two operand instruction where any addressing mode may be used to specify both the operands is the MOVE instruction. 12 Instructions (contd..) Consider the instruction: ADD #9, D3 Add 9 to contents of D3 and store result in D3. 15 12 11 9 8 7 6 5 1 1 01 dst 0 0 src size operation Binary Hex 1 1 0 10 1 1 00 1 1 11 1 0 0 D 6 7 C •Bits 15-12 and 8 specify the operation. •Bits 9-11 specify the destination register D3. •Bits 7-5 specify the size of the operands, default is word, with a bit pattern 01. •Bits 5-0 identifies that the source operand is immediate with the pattern 111100. •The actual value of the source operand is in the extension word. •Binary and hex encoding of the OP-code word are shown. 13 Instructions (contd..) Consider the instruction: ADD #9, D3 Add 9 to contents of D3 and store result in D3. i D67C i +2 OP-code word 9 Immediate operand i +4 D3 25 D3 34 PC i PC i+4 Before instruction fetch After instruction execution •OP-code word is at location i. •Operand 9 is in the extension word at location i+2. •After the OP-code word is fetched PC points to i+2. •After the extension word is fetched, PC points to i+4. •Contents of D3 before execution are 25, and after execution are 34. 14 Assembly language Instructions can operate on operands of three sizes, size of the operands must be indicated: Size of the operand is indicated in the operation mnemonic, using L for long-word, W for word, and B for byte. ADD.L indicates ADD operation with operand size long-word. Representation of numbers: Default representation is in the decimal form. Prefix $ is used for hexadecimal, and % for binary. Representation of Alphanumeric characters: Placed within quotes, replaced by the assembler with their ASCII codes. Several characters may be placed in a string between quotes. 15 Assembler directives ORG: Starting address of an instruction or data block. EQU: Equating names with numerical values DC (Define Constant): Data constants. Size indicator may be appended to specify the size of the data constants. DC.B, DC.W, DC.L. DS (Define Storage): Reserve a block of memory. Names can be associated with memory locations using labels as described earlier. 16 Condition code flags Five condition code flags, stored in the status register. N Z V C X – Extend flag, set the same way as C flag, but not affected by many instructions. Useful for precision arithmetic. Since operands can be of three sizes (byte, word and longword): C and X flags depend on the carry-out from bit positions 7, 15 and 31. 17 Branch instructions Recall that a conditional branch instruction causes program execution to continue with the instruction at the branch target address if the condition is met, else the instruction immediately following the branch is executed. Branch target address is computed by adding an offset to the PC. Branch instructions with two types of offset, 8 and 16 bit. Branch instructions with 8-bit offset: Offset is included in the OP-code word. Branch target address is within –127 to 128 of the value in PC. PC contents are incremented at each fetch, offset defines the distance from the word that follows the branch instruction OPcode word. 18 Branch instructions (contd..) Branch instructions with a 16-bit offset: Offset is specified in an extension word following the OP-code word. Branch target address can be in the range of +32K to –32K from the value in PC. Offset is the distance from the extension word to the branch target. 16 conditional branch instructions, each with two offsets. 1 unconditional branch instruction. 19 Stacks and subroutines Stack can be implemented using any of the address registers, with the help of autoincrement and autodecrement addressing modes. Register A7 is designated as the processor stack pointer, and it points to the processor stack. Stack used by the processor for all operations that it performs automatically, such as subroutine linkage. Two different stack pointer registers, for two modes of operation: Supervisor mode: All machine instructions can be executed. User mode: Certain privileged instructions cannot be executed. Application programs: user mode. System software: supervisor mode. 20 Stacks and subroutines (contd..) Stack may be used for: Subroutine linkage mechanism. Passing parameters to a subroutine and returning result from the subroutine. Branch-to-Subroutine (BSR) is used to call a subroutine: Causes the contents of the PC to be pushed onto the stack. Return from Subroutine (RTS) is used to return from a subroutine: Causes the return address at the top of the stack to be popped into the PC. 21 Simple programs in 68000 assembly language. Add the contents of memory locations A and B, and place the result in C. Size of operands A and B is word. Recall that: - For a two-operand instruction, one of the operands has to be placed in a data register D0 through D7. MOVE A, D0 ADD B, D0 MOVE D0, C Move A to register D0. Add the contents of location B to D0 and store the result in D0. Transfer the result in register D to location C. 22 Simple programs in 68000 assembly language. A = 201150 63910 B = 201152 - 21510 201200 OP-code word 201202 20 201204 1150 201206 OP-code word 201208 20 20120A 1152 20120C OP-code word 20120E 20 201210 2200 •Add the contents of A and B, and place the result in location MOVE A,D0 C. •Size of the operands at locations A and B is 16 bits. •A, B, and C and the program ADD B,D0 to add is stored in the memory MOVE D0,C C = 202200 After execution, [202200] = 424 23 Simple programs in 68000 assembly language Program to add A and B, and store the result in C with assembler directives. Memory address label Assemblerdirectives C A B Statemen ts that generatemachine instructions Assemblerdirective Operation EQU ORG DC.W DC.W ORG MOVE ADD MOVE END Addressing or data information $202200 $201150 639 – 215 $201200 A,D0 B,D0 D0,C 24 Simple programs in 68000 assembly language. Add N numbers: - The first number is stored at the starting address NUM1. - The count of entries to be added is stored at address N. - Store the result at location SUM. - Size of each number to be added is Word. LOOP MOVE.L N,D1 MOVEA.L #NUM1,A2 CLR.L ADD.W SUBQ.L BGT MOVE.L D0 (A2)+,D0 #1,D1 LOOP D0,SUM N contains n, the number of entries to be added, and D1 is usedas a counter that determineshow many times to executethe loop. A2 is usedas a pointer to the list entries. It is initialized to NUM1, the address of the first entry . D0 is used to accumulate the sum. Successiv e numbers are added in D0. Decremen t the counter. If [D1]> 0, executethe loop again. Store the sum in SUM. 25