1 / 31

Mastery of MIPS Assembly Language - Fibonacci Procedure Explanation

Dive into the Fibonacci procedure in MIPS assembly language, learning how to calculate Fibonacci numbers recursively and efficiently. Understand register usage and the stack functionality for proper execution.

paul2
Télécharger la présentation

Mastery of MIPS Assembly Language - Fibonacci Procedure Explanation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Assembly Language IICPSC 321 Andreas Klappenecker

  2. Any Questions?

  3. Administrative Issues • Confusion about Section 502 • Your lab assignments should be turned in during your lab. • Take advantage of the Teaching Assistant and the Peer Teacher. • Learn about many computer related topics during the workshops offered by our graduate students. • Go to seminars to learn about current trends in computer science.

  4. The Story so far… • We introduced numerous MIPS assembly language instructions. • We are now familiar with registers and register usage conventions. • We know how to use system calls, basic I/O • We have learned how the stack works • What is missing? Practice! Practice! Practice!

  5. Stack $sp = $sp - 12 The stack pointer is contained in register $sp. The stack grows from above. If you want to push a word onto the stack: $sp = $sp – 4 Efficiency: If you want to push 3 registers onto the stack, subtract 12! high address stack pointer$sp -> low address

  6. Fibonacci Procedure • We want to write a recursive procedure fib that performs the following calculation: • If n=0 or n=1, then f(0)=0 and f(1)=1 • simply return the argument • If n>1, then f(n)=f(n-1)+f(n-2) • recursively call fib

  7. Top Down Design <<Fibonacci>>= fib: <<save registers>> <<calculation>> <<restore registers>> We assume that $a0 contains the argument n, register $v0 the result, and $s0 some intermediate result.

  8. Calculation <<calculation>>= bgt $a0,1, gen # if n>1, goto generic case move $v0,$a0 # output = input if n=0 or n=1 j rreg # goto restore registers gen: sub $a0,$a0,1 # param = n-1 jal fib # compute fib(n-1) move $s0,$v0 # save fib(n-1) sub $a0,$a0,1 # set param to n-2 jal fib # and make recursive call add $v0, $v0, $s0 # $v0 = fib(n-2)+fib(n-1)

  9. <<save registers>>= subi $sp,$sp,12 sw $a0, 0($sp) sw $s0, 4($sp) sw $ra, 8($sp) <<restore registers>>= lw $a0, 0($sp) lw $s0, 4($sp) lw $ra, 8($sp) addi $sp,$sp, 12 jr $ra Save and Restore Registers

  10. fib: sub $sp,$sp,12 # save registers on stack sw $a0, 0($sp) # save $a0 = n sw $s0, 4($sp) # save $s0 sw $ra, 8($sp) # save return address $ra bgt $a0,1, gen # if n>1 then goto generic case move $v0,$a0 # output = input if n=0 or n=1 j rreg # goto restore registers gen: sub $a0,$a0,1 # param = n-1 jal fib # compute fib(n-1) move $s0,$v0 # save fib(n-1) sub $a0,$a0,1 # set param to n-2 jal fib # and make recursive call add $v0, $v0, $s0 # $v0 = fib(n-2)+fib(n-1) rreg: lw $a0, 0($sp) # restore registers from stack lw $s0, 4($sp) # lw $ra, 8($sp) # add $sp, $sp, 12 # decrease the stack size jr $ra

  11. Quiz • How do you calculate f(12) with the Fibonacci procedure that we have written? li $a0, 12 jal fib

  12. Short Discussion • You are now familiar with the basics of the MIPS assembly language. • Start experimenting! Do it yourself! • What kind of techniques do you use to design your programs? • What style of documentation are you using? • Are you familiar with TeX or LaTeX?

  13. What Next? • We need a more detailed knowledge about the instruction formats to fully appreciate certain restrictions. • The functional interface is easy to understand, since it is basically familiar procedural programming • We need to understand how the computer interprets the instruction, so that we can transition to the discussion of the MIPS hardware architecture

  14. Machine Language What does that mean? • Machine language level programming means that we have to provide the bit encodings for the instructions • For example, add $t0, $s1, $s2 represents the 32bit string • 00000010001100100100000000100000 • Assembly language mnemonics usually translate into one instruction • We also have pseudo-instructions that translate into several instructions

  15. Instruction Word Formats Register format Immediate format Jump format op-code rs rt rd shamt funct 6 5 5 5 5 6 op-code rs rt immediate value 6 5 5 16 op-code 26 bit current segment address 6 26

  16. Register Format (R-Format) Register format op: basic operation of instruction funct: variant of instruction rs: first register source operand rt: second register source operand rd: register destination operand shamt: shift amount op-code rs rt rd shamt funct 6 5 5 5 5 6

  17. Watson, the case is clear… • add $t0, $s1, $s2 • 00000010001100100100000000100000 • 000000 10001 10010 01000 00000 100000 • Operation and function field tell the computer to perform an addition • 000000 10001 10010 01000 00000 100000 • registers $17, $18 and $8 op-code rs rt rd shamt funct 6 5 5 5 5 6

  18. Number Value Name Registers return values from functions pass parameters to functions $t0-$t7 are caller saved registers – use these registers in functions $s0-$s7 are callee-saved registers – use these registers for values that must be maintained across function calls.

  19. Watson, the case is clear… • add $t0, $s1, $s2 • 00000010001100100100000000100000 • 000000 10001 10010 01000 00000 100000 • source registers $s1=$17 and $s2=$18 and target register $t0=$8 op-code rs rt rd shamt funct 6 5 5 5 5 6

  20. R-Format Example Register format (op, funct)=(0,32): add rs=17: first source operand is $s1 rt=18: second source operand is $s2 Rd=8: register destination is $t0 add $t0, $s1, $s2 0 17 18 8 0 32 6 5 5 5 5 6

  21. Immediate Format (I-Format) Immediate format op determines the instruction (op <> 0) rs is the source register rt is the destination register 16bit immediate value op rs rt immediate value 6 5 5 16

  22. I-Format Example Immediate format op=8 means addi rs=29 means source register is $sp rt=29 means $sp is destination register immediate value = 4 addi $sp, $sp, 4 8 29 29 4 6 5 5 16

  23. Problem • The MIPS assembly language has the command andi, an immediate bit-wise and operation • We can say li $s0, 0xCDEF1234 to load register $s0 with the content 0xCDEF1234 • Why is this strange? • In the immediate format, you can only load 16 bits, but the constant is 32 bits!

  24. Pseudo-Instructions • li $s0, 0xCDEF1234 is a pseudo-instruction • It is a convenient shorthand for lui $at, 0xCDEF ori $s0, $at, 0x1234 • The register $at is used here by the assembler; this is the reason why you should not use this register.

  25. Puzzle • How can we swap the content of two registers, say $s0 and $s1, without accessing other registers or memory? • Solution: xor $s0, $s0, $s1 xor $s1, $s0, $s1 xor $s0, $s0, $s1

  26. MIPS Addressing Modes • Immediate addressing • Register addressing • Base displacement addressing • PC-relative addressing • address is the sum of the PC and a constant in the instruction • Pseudo-direct addressing • jump address is 26bits of instruction concatenated with upper bits of PC

  27. Addressing Modes • Register Addressing • add $s1, $s2, $s3 • $s1 = $s2 + $s3 • Immediate Addressing • addi $s1, $s2, 100 • $s1 = $s2 + 100

  28. Addressing Modes • Base addressing • lw $s1, 100($s2) • $s1 = Memory[$s2+100] • PC-relative branch • beq $s1, $s2, 25 • if ($s1 == $s2) goto PC + 4 + 100

  29. Addressing Modes • Pseudo-direct addressing • j 1000 • goto 1000 • concatenate 26bit address with upper bits of the PC • Study section 3.8 for further details • In particular, get used to Figure 3.18

  30. Conclusion • Read Chapter 3 in conjunction with Appendix A • You will miss many details – go back and read it again and again after we have discussed the MIPS hardware architecture. • Now is the time to get familiar with assembly language programming! No cheating! • Make sure that you program at least two procedures this weekend. • Have a close look at the simulator. What does it do with pseudo-instructions?

More Related