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Assembly Language

This chapter provides an introduction to assembly language programming, including the components of a computer, MIPS registers, MIPS assembly instructions, instruction representation examples, procedure calls, memory allocation, and more.

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Assembly Language

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  1. Assembly Language Chapter 3

  2. Computer Languages • High-level languages • Java, Python, C, ASP, HTML, … • Low-level languages • Assembly language: symbolic representation of machine instructions. • Assembler: a compiler which translates from an assembly languageto a machine language. • Machine languages Chapter 3 Assembly Languages

  3. Components of a computer: More detail processor memory registers ALU data program Instruction register Program counter data address Control units Chapter 3 Assembly Languages

  4. Applications Operating systems Instruction set Functional units Finite state machine Logic gates Electronics Levels of descriptions of computer systems • computer architecture begins at the instruction set. • An instruction set is what a programmer at the lowest level sees of a processor • one instruction set with different level of performance for many models, based on the implementation of a control unit via “microprogram”. • Present day processor designs converge, their instruction sets become more similar than different. Chapter 3 Assembly Languages

  5. Registers NameReg. No. Usage $zero 0 hardwired 0 $v0-$v1 2-3 return value and expression evaluation $a0-$a3 4-7 arguments $t0-$t7 8-15 temporary values $s0-$s7 16-23 saved values $t8-$t9 24-25 more temporary values $gp 28 global pointer $sp 29 stack pointer $fp 30 frame pointer $ra 31 return address Chapter 3 Assembly Languages

  6. MIPS operands Chapter 3 Assembly Languages

  7. MIPS Assembly Instructions: Arithmetic Chapter 3 Assembly Languages

  8. MIPS Assembly Instructions: Data Transfer Chapter 3 Assembly Languages

  9. MIPS Assembly Instructions: Logical Chapter 3 Assembly Languages

  10. MIPS Assembly Instructions Chapter 3 Assembly Languages

  11. MIPS Assembly Instructions: Conditionals Chapter 3 Assembly Languages

  12. x = y+z+5; w = z-y; lw $t0, 18($zero) # load y lw $t1, 22($zero) # load z add $t2, $t0, $t1 # y+z addi $t2, $t2, 5 # (y+z)+5 sw $t2, 14($zero) # store x sub $t3, $t1, $t2 # z-y sw $t3, 10($zero) # store w if (a<b) a++; else b++; lw $t0, 50($zero) # load a lw $t1, 54($zero) # load b slt $t2, $t0, $t1 # a<b beq $t2, $zero, else # if addi $t0, $t0, 1 # a++ sw $t0, 50($zero) # store a else: addi $t1, $t1, 1 # b++ sw $t1, 54($zero) # store b Example 10 14 18 22 50 54 Chapter 3 Assembly Languages

  13. Basic instruction formats R- format I- format J- format Chapter 3 Assembly Languages

  14. Instruction Representation: Examples Chapter 3 Assembly Languages

  15. Instruction Representation: Examples add $t0, $t1, $t2 Chapter 3 Assembly Languages

  16. Instruction Representation: Examples sub $s0, $s1, $s2 Chapter 3 Assembly Languages

  17. Instruction Representation: Examples addi $s0, $s1, 100 Chapter 3 Assembly Languages

  18. Instruction Representation: Examples j 10000 Chapter 3 Assembly Languages

  19. Instruction Representation: Examples beq $s0, $s1, 100 Chapter 3 Assembly Languages

  20. MIPS Simulator http://pages.cs.wisc.edu/~larus/spim.html Chapter 3 Assembly Languages

  21. Case/ Switch statement switch (i) { case 0: i++; break; case 1: j++; break; case 4: k++; } • Jump (address) table • Table storing addresses of different cases Chapter 3 Assembly Languages

  22. switch (i) { case 0: i++; break; case 1: j++; break; case 4: k++; } # variable i is in $t1, const 5 is in $t0 # start addr of jump table (e.g. 3010) is in $s0 blt $t1, $zero, out bge$t1, $t0, out # multiply i by 4 for word size sll $t1, $t1, 2 # find entry in the jump table add $t1, $t1, $s0 # load addr in jump table to $t2 lw $t2, 0($t1) jr $t2 L2010: … j out L20485: … out: Case statement and jrinstruction Chapter 3 Assembly Languages

  23. Procedure calls • Steps to Execute procedure • place parameters in a place where procedure can access them • transfer control to the procedure • acquire the storage needed for the procedure • perform the desired task • place the result value in a place where the calling program can access it • return to the control point of origin Chapter 3 Assembly Languages

  24. Memory allocation for programs $sp Stack Heap (dynamic data) Static data Code (text segment) Reserved $fp $gp pc Chapter 3 Assembly Languages

  25. Registers • $a0-$a3 • four argument registers in which to pass parameters • $v0-$v1 • two value registers in which to return values • $ra • one return address register to return to the point of origin Chapter 3 Assembly Languages

  26. jal jump and link • jumps to an address and simultaneously saves the address of the following instruction in register $ra • jal ProcedureAddress • jal puts PC + 4 in $ra • to return, after completion of the procedure: • jr $ra • calling program (caller) puts parameter values in $a0-$a3 • callee performs calculations and places result in $v0-$v1 Chapter 3 Assembly Languages

  27. Call sequence • Put arguments in $a0-$a3 • The rest of the arguments (if exist) are placed in the frame or activation record • Allocate a frame in stack (update $sp) • Save values of $s0-$s7, $fp, $rain the frame • Update $fp • Execute jal instruction Chapter 3 Assembly Languages

  28. Return sequence • Put the return value in $v0 • Restore registers $s0-$s7 • Restore $fp • Pop the frame from stack • Execute jr $ra Chapter 3 Assembly Languages

  29. int leaf_example (int g, int h, int i, int j) { int f; f = (g + h) - (i + j); return f; } leaf example: addi $sp, $sp, -12 #adjust stack to make room for 3 items sw $t1, 8($sp) #save register $t1 for use afterwards sw $t0, 4($sp) #save register $t0 for use afterwards sw $s0, 0($sp) #save register $s0 for use afterwards add $t0, $a0, $a1 #register $t0 contains g + h add $t1, $a2, $a3 #register $t1 contains i + j sub $s0, $t0, $t1 #f = $t0 - $t1, which is (g+h)-(i+j) add $v0, $s0, $zero #returns f ($v0 = $s0 + 0) lw $s0, 0($sp) #restore register $s0 for caller lw $t0, 4($sp) #restore register $t0 for caller lw $t1, 8($sp) #restore register $t1 for caller addi $sp, $sp, 12 #adjust stack to delete 3 items jr $ra #jump back to calling routing Example 1 Chapter 3 Assembly Languages

  30. int fact (int n) { if (n<2) return(1); else return(n*fact(n-1); } fact: addi $sp, $sp, -8 #adjust stack to make room for 2 items sw $ra, 4($sp) #save the return address sw $a0, 0($sp) #save the argument n slti $t0, $a0, 2 #test for n<2 beq $t0, $zero, L1 #if n>=2 goto L1 addi $sp, $sp, 8 #adjust stack to delete 2 items addi $v0, $zero, 1 #else return 1 jr $ra #jump back to calling routing L1: addi $a0, $a0, -1 #if n>=1 calculate n-1 jal fact #call fact with n-1 lw $a0, 0($sp) #return from jal, restore argument lw $ra, 4($sp) #restore return address addi $sp, $sp, 8 #adjust stack to delete 2 items mul $v0, $a0, $v0 # return n* fact(n-1) jr $ra #jump back to calling routing Example 2 Chapter 3 Assembly Languages

  31. fact: addi $sp, $sp, -8 sw $ra, 4($sp) sw $a0, 0($sp) slti $t0, $a0, 2 beq $t0, $zero, L1 addi $sp, $sp, 8 addi $v0, $zero, 1 jr $ra L1: addi $a0, $a0, -1 jal fact lw $a0, 0($sp) lw $ra, 4($sp) addi $sp, $sp, 8 mul $v0, $a0, $v0 jr $ra int fact (int n) { if (n<2) return(1); else return(n*fact(n-1); } … fact(2); Execution example Push frame Call fact (2)addi $a0, $zero, 2 (03A0)jal fact $a0 2 n<2 ret 1 1 stack 03A4 ($ra) 2($a0) $ra 03A4 0344 (0340) 0344 ($ra) 1($a0) restore $v0 1 2 n>=2 ret n*(n-1)! Chapter 3 Assembly Languages

  32. 32-bit constants • A register can contain an integer between -2147483648 (-231) and 2147483647 • Immediate data can be between -32768 (-215) and 32767 • addi $t0, $zero, 32767 • How can we use 32-bit constant? • lui (load upper immediate) instruction Chapter 3 Assembly Languages

  33. load upper immediate instruction: lui • Store 16-bit immediate data in the upper half of the register lui $t0, 16384# put 0100 0000 0000 0000 in the upper half of $t0 • To store a 32-bit constant in a register add $t0, $zero, $zero lui $t0, 16384 addi $t0, $t0, 5435 0100 0000 0000 0000 Chapter 3 Assembly Languages

  34. Jumping to far-away instruction • In beq or bne instructions, only  215 distance can be specified. • To jump to further instructions beq $t0, $t1, L1 ... L1: j L2 ... L2: Chapter 3 Assembly Languages

  35. Memory Addressing Modes • direct • mem[address] • register indirect • mem[content[reg]] • mem[content[address]] • register indirect displacement • mem[content[reg]+displacement] • mem[content[address]+displacement] • register indirect index and displacement Chapter 3 Assembly Languages

  36. MIPS Addressing Modes • Register addressing (R-format) • add $t0,$t1,$t2 • Base (Displacement) addressing (I-format) • lw $t1, 12($t2) • Immediate addressing (J-format) • addi $t0, $t1, 34 register memory + register Chapter 3 Assembly Languages

  37. MIPS Addressing Modes • PC-relative addressing (I-format) • beq $t0, $t1, label • Pseudo-direct addressing • j label memory + PC : PC Chapter 3 Assembly Languages

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