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Assembly Language II CPSC 321 Andreas Klappenecker.

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Presentation on theme: "Assembly Language II CPSC 321 Andreas Klappenecker."— Presentation transcript:

1 Assembly Language II CPSC 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 8($sp) 4($sp) 0($sp) high address low address stack pointer $sp -> 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! $sp = $sp - 12

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 >= fib: > > We assume that $a0 contains the argument n, register $v0 the result, and $s0 some intermediate result.

8 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 and Restore Registers >= subi $sp,$sp,12 sw $a0, 0($sp) sw $s0, 4($sp) sw $ra, 8($sp) >= lw $a0, 0($sp) lw $s0, 4($sp) lw $ra, 8($sp) addi $sp,$sp, 12 jr $ra

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 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 What does that mean?

15 Instruction Word Formats Register format Immediate format Jump format op-code rs rt rd shamt funct op-code rs rt immediate value op-code 26 bit current segment address 6 55 16 6 555 5 6 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 6 555 5 6

17 Watson, the case is clear… add $t0, $s1, $s2 00000010001100100100000000100000 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 6 555 5 6

18 0$zero 1$at 2$v0 3$v1 4$a0 5$a1 6$a2 7$a3 8$t0 9$t1 10$t2 11$t3 12$t4 13$t5 14$t6 15$t7 16$s0 17$s1 18$s2 19$s3 20$s4 21$s5 22$s6 23$s7 24$t8 NumberNameValue Registers pass parameters to functions return values from functions $s0-$s7 are callee-saved registers – use these registers for values that must be maintained across function calls. $t0-$t7 are caller saved registers – use these registers in functions

19 Watson, the case is clear… add $t0, $s1, $s2 00000010001100100100000000100000 source registers $s1=$17 and $s2=$18 and target register $t0=$8 op-code rs rt rd shamt funct 6 555 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 6 555 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 55 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 55 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


28 Addressing Modes Register Addressing add $s1, $s2, $s3 $s1 = $s2 + $s3 Immediate Addressing addi $s1, $s2, 100 $s1 = $s2 + 100

29 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

30 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

31 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?

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