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1 CS/COE0447 Computer Organization & Assembly Language Chapter 2 Part 4.

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Presentation on theme: "1 CS/COE0447 Computer Organization & Assembly Language Chapter 2 Part 4."— Presentation transcript:

1 1 CS/COE0447 Computer Organization & Assembly Language Chapter 2 Part 4

2 2 Topics Creating executables –Assembler, linker, loader –Compilers versus interpreters What does the assembler do for you? Review of branch and jump instructions

3 3 “C Program” Down to “Numbers” swap: muli$2, $5, 4 add$2, $4, $2 lw$15, 0($2) lw$16, 4($2) sw$16, 0($2) sw$15, 4($2) jr$31 void swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; } 00000000101000010… 00000000000110000… 10001100011000100… 10001100111100100… 10101100111100100… 10101100011000100… 00000011111000000… compiler assembler

4 4 To Produce an Executable source file.asm/.s source file.asm/.s source file.asm/.s object file.obj/.o object file.obj/.o object file.obj/.o library.lib/.a executable.exe assembler linker

5 5 Linker

6 6 An Object File Header –Size and position of other pieces of the file Text segment –Machine codes Data segment –Binary representation of the data in the source Relocation information –Identifies instructions and data words that depend on absolute addresses Symbol table –Keeps addresses of global labels –Lists unresolved references Debugging information –Contains a concise description of the way in which the program was compiled

7 7 An Assembler Expands macros –Macro is a sequence of operations conveniently defined by a user –A single macro can expand to many instructions Determines addresses and translates source into binary numbers –Record in “symbol table” addresses of labels –Resolve branch targets and complete branch instructions’ encoding –Record instructions that need be fixed after linkage Packs everything in an object file “Two-pass assembler” –To handle forward references –bne $t0,$t1,next_label –next_label:

8 8 Assembler Directives Guides the assembler to properly handle following codes with certain considerations.text –Tells assembler that codes follow.data –Tells assembler that data follow.align –Directs aligning the following items.global –Tells to treat the following symbol as global.asciiz –Tells to handle the following as a “string”

9 9 Macro Example.data int_str:.asciiz“%d”.text.macroprint_int($arg) la $a0, int_str mov $a1, $arg jal printf.end_macro … print_int($7) la $a0, int_str mov $a1, $7 jal printf

10 10 Branch and Jump Instructions Review

11 11 Instruction Format beq, bne PC = PC + 4 + BranchAddr BranchAddr = {14{immediate[15]},immediate,2'b0} Two issues: –Assembly  machine code –Execution of the machine code 16-bit immediatertrsop I

12 12 bne $t0,$s5,exitloop addi $s3,$s3,1 j loop exitloop: add $s6,$s3,$zero 0x00400024 0x00400028 0x0040002c 0x00400030 BNE machine code in binary: 000101 01000 10101 0000000000000010 BNE machine code in hex: 15150002 When BNE instruction is executed (condition true): Next address = PC + 4 + BranchAddr Next address = 00400024 + 4 + 00000008 = 00400030 … address of the exitloop inst! BranchAddr = {14{immediate[15]},immediate,2'b0} = {00000000000000,0000000000000010,00} = 0000 0000 0000 00 00 0000 0000 0000 1000 = 0x00000008

13 13 BranchAddr: Why 2’b0 at the end? BranchAddr might have been: {number,00} = number * 4 (like shifting left by 2) Recall: the immediate field of the instruction contains the number of instructions away the label is Each instruction is 4 bytes, so multiplying by 4 gives you the number of bytes away it is. This is the number to add to PC + 4 to jump to the label. {16{immediate[15]},immediate}

14 14 BranchAddress: Why 2’b0 at the end? If immediate instead were the number of bytes away the label is, then we would be wasting 2 bits Since all instruction addresses are multiples of 4, the bottom 2 bits are always 00 By not including those bits in the immediate field of the machine code, branch instructions can be used to jump 4 times further away

15 15 Instruction Format, cont’d The address of next instruction is obtained by concatenating with PC PC = {PC[31:28],address,2’b0} 26-bit addressopJump

16 16 0x00400018 bne $s4, $s5, ELSE 0x0040001c add $s3, $s2, $s5 0x00400020 j EXIT ELSE: 0x00400024 sub $s3, $s2, $s5 0x00400028 addi $s5, $s5, 1 0x0040002c EXIT: addi $s4,$s4,1 j instruction machine code: Hex: 0810000b Look at execution: PC = {PC[31:28],address,00} PC[31:28] = 0000 address = 00 0001 0000 0000 0000 0000 1011 {0000, address, 00} = 0000 00 0001 0000 0000 0000 0000 1011 00 BIN 0 0 4 0 0 0 2 c HEX The address EXIT stands for!


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