Csci 136 Computer Architecture II – Summary of MIPS ISA

1. Csci 136 Computer Architecture II– Summary of MIPS ISA Xiuzhen Cheng cheng@gwu.edu

2. Announcement • Homework assignments #2: • Readings: Sections 3.6, 3.7, 3.10, 3.11, 3.13, 3.14 • Problems 3.5, 3.10, 3.12, 3.22, 3.23, 3.24, 3.25 • Project #1 is due on 11:59PM, Feb 13, 2003.

3. What is an ISA? • A very important abstraction • Provide the interface between the low-level software and hardware • May have multiple hardware implementations • Needs to answer the following questions • What is the minimum instruction set to be supported? • Use general purpose register or not? • CIRS or RISC design? • Instruction format? • Addressing mode? • … …

4. Register Register (register-memory) (load-store) Load R1,A Add R1,B Store C, R1 Basic ISA Classes • Accumulator Architecture • Stack Architecture • Load-Store (General Purpose Register) Architecture • Register – Register • Register – Memory • Memory – Memory • Code sequences for (C=A+B) for 4 classes of instruction set Stack Accumulator Push A Load A Load R1,A Push B Add B Load R2,B Add Store C Add R3,R1,R2 Pop C Store C,R3

5. ° Since 1975, all machines use general purpose registers ° Advantages of registers • registers are faster than memory • registers are easier for a compiler to use e.g., (A*B) – (C*D) – (E*F) can do multiplies in any order - vs. stack • registers can hold variables - memory traffic is reduced, so program is sped up (since registers are faster than memory) code density improves (since register named with fewer bits - than memory location) General Purpose Register Dominates

6. Memory Addressing ° Since 1980 almost every machine uses addresses to level of 8-bits (byte) ° 2 questions for design of ISA: • Since one could read a 32-bit word as four loads of bytes from sequential byte addresses or as one load word from a single byte address, How do byte addresses map onto words? Can a word be placed on any byte boundary? • What about MIPS?

7. ISA Operation Summary • Support these simple instructions, since they will dominate the number of instructions executed: • load, • store, • add, • subtract, • move register-register, • and, • shift, • compare equal, compare not equal, • branch, jump, • call, • return;

8. Summary: Salient features of MIPS • 32-bit fixed format inst (3 formats) • 32 32-bit GPR (R0 contains zero) and 32 FP registers (and HI LO) • partitioned by software convention • 3-address, reg-reg arithmetic instr. • Single addressing mode for load/store: base+displacement • no indirection, scaled • 16-bit immediate plus LUI • Simple branch conditions • compare against zero or two registers for =, • no integer condition codes

9. Summary: MIPS Instruction set design • Use general purpose registers with a load-store architecture: yes or no? • Provide 32 general purpose registers plus separate floating-point registers: • What’s the addressing mode supported by MIPS? • Use fixed instruction encoding if interested in performance and use variable instruction encoding if interested in code size :

10. Summary: MIPS Instruction set design • Support these data sizes and types: 8-bit, 16-bit, 32-bit integers and 32-bit and 64-bit IEEE 754 floating point numbers: • Support these simple instructions, since they will dominate the number of instructions executed: load, store, add, subtract, move register-register, and, shift, compare equal, compare not equal, branch, jump, call, and return: • Aim for a minimalist instruction set:

11. MIPS Hardware Design Principles • Simplicity favors regularity • Keeping the hardware simple! • R-Type instruction format • Smaller is faster • 32 general purpose registers, no more, no less. • Good design demands good compromises • R, I, J, 3 types of instruction formats • Make the common case fast! • I-type instructions for constant numbers

12. In-Class Exercise: Reverse a String • Write a MIPS procedure to reverse a null-terminated character string. Assume the address of the string is in $a0 and the address of the reversed string is in $a1. Also assume the spaces needed by the reversed string have been pre-allocated.

13. In-Class Exercise: Reverse a String • Write a MIPS procedure to reverse a null-terminated character string. Assume the address of the string is in $a0 and the address of the reversed string is in $a1. Also assume the spaces needed by the reversed string have been pre-allocated. reverseStr: addiu $sp, $sp, -32 sw $s0, 16($sp) sw $s1, 20($sp) move $s0, $a0 move $s1, $a1 addi $sp, $sp, -1 sb $zero, 0($sp) push: lbu $t0, 0($s0) beq $t0, $zero, pop addi $s0, $s0, 1 addi $sp, $sp, -1 sb $t0, 0($sp) j push pop: lbu $t0, 0($sp) addi $sp, $sp, 1 sb $t0, 0($s1) beq $t0, $zero, done addi $s1, $s1, 1 j pop done: lw $s0, 16($sp) lw $s1, 20($sp) addi $sp, $sp, 32 jr $ra

14. Questions?