machine-level programming 3 control flow

19
Machine-Level Programming 3 Control Flow Topics Control Flow Switch Statements Jump Tables

Upload: john-mckee

Post on 31-Dec-2015

20 views

Category:

Documents


0 download

DESCRIPTION

Machine-Level Programming 3 Control Flow. Topics Control Flow Switch Statements Jump Tables. Switch Statements. Implementation Options Jump Table Lookup branch target Constant time Possible when cases are small integer constants Series of conditionals Organize in tree structure - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Machine-Level Programming 3 Control Flow

Machine-Level Programming 3Control Flow

Topics Control Flow

Switch StatementsJump Tables

Page 2: Machine-Level Programming 3 Control Flow

– 2 – CMSC 313, F’09

Switch Statements

Implementation Options Jump Table

Lookup branch targetConstant timePossible when cases are small integer constants

Series of conditionalsOrganize in tree structureLogarithmic performance

GCCPicks one based on case structure

Page 3: Machine-Level Programming 3 Control Flow

– 3 – CMSC 313, F’09

Switch Statement Example

Features Multiple case labels Fall through cases Missing cases

long switch_eg (long x, long y, long z){ long w = 1; switch(x) { case 1: w = y*z; break; case 2: w = y/z; /* Fall Through */ case 3: w += z; break; case 5: case 6: w -= z; break; default: w = 2; } return w;}

Page 4: Machine-Level Programming 3 Control Flow

– 4 – CMSC 313, F’09

Jump Table Structure

Code Block0

Targ0:

Code Block1

Targ1:

Code Block2

Targ2:

Code Blockn–1

Targn-1:

•••

Targ0

Targ1

Targ2

Targn-1

•••

jtab:

target = JTab[x];goto *target;

switch(x) { case val_0: Block 0 case val_1: Block 1 • • • case val_n-1: Block n–1}

Switch Form

Approx. Translation

Jump Table Jump Targets

Page 5: Machine-Level Programming 3 Control Flow

– 5 – CMSC 313, F’09

Switch Statement Example (IA32)

Setup: switch_eg:pushl %ebp # Setupmovl %esp, %ebp # Setuppushl %ebx # Setupmovl $1, %ebx # w = 1movl 8(%ebp), %edx # edx = xmovl 16(%ebp), %ecx # ecx = zcmpl $6, %edx # x:6ja .L61 # if > goto defaultjmp *.L62(,%edx,4) # goto JTab[x]

long switch_eg (long x, long y, long z){ long w = 1; switch(x) { . . . } return w;}

Page 6: Machine-Level Programming 3 Control Flow

– 6 – CMSC 313, F’09

Assembly Setup Explanation

Table Structure Each target requires 4 bytes Table base address at .L62

JumpingDirect Jumpjmp .L61 Jump target is denoted by label .L61

Indirect Jumpjmp *.L62(,%edx,4) Start of jump table denoted by label .L62 Register %edx holds x Must scale by factor of 4 to get offset into table Fetch target from effective Address .L61 + x*4

Only for 0 x 6

Page 7: Machine-Level Programming 3 Control Flow

– 7 – CMSC 313, F’09

Jump Table

.section .rodata .align 4.L62:.long .L61 # x = 0.long .L56 # x = 1.long .L57 # x = 2.long .L58 # x = 3.long .L61 # x = 4.long .L60 # x = 5.long .L60 # x = 6

Table Contents

switch(x) { case 1: // .L56 w = y*z; break; case 2: // .L57 w = y/z; /* Fall Through */ case 3: // .L58 w += z; break; case 5: case 6: // .L60 w -= z; break; default: // .L61 w = 2; }

Page 8: Machine-Level Programming 3 Control Flow

– 8 – CMSC 313, F’09

Code Blocks (Partial).L61: // Default case

movl $2, %ebx # w = 2movl %ebx, %eax # Return wpopl %ebxleaveret

.L57: // Case 2:movl 12(%ebp), %eax # ycltd # Div prepidivl %ecx # y/z movl %eax, %ebx # w = y/z

# Fall through.L58: // Case 3:

addl %ecx, %ebx # w+= zmovl %ebx, %eax # Return wpopl %ebxleaveret

switch(x) { . . . case 2: // .L57 w = y/z; /* Fall Through */ case 3: // .L58 w += z; break; . . . default: // .L61 w = 2; }

Page 9: Machine-Level Programming 3 Control Flow

– 9 – CMSC 313, F’09

Code Blocks (Rest).L60: // Cases 5&6:

subl %ecx, %ebx # w –= zmovl %ebx, %eax # Return wpopl %ebxleaveret

.L56: // Case 1:movl 12(%ebp), %ebx # w = yimull %ecx, %ebx # w*= zmovl %ebx, %eax # Return wpopl %ebxleaveret

switch(x) { case 1: // .L56 w = y*z; break; . . . case 5: case 6: // .L60 w -= z; break; . . . }

Page 10: Machine-Level Programming 3 Control Flow

– 10 – CMSC 313, F’09

IA32 Object CodeContext

Label .L61 becomes address 0x8048630 Label .L62 becomes address 0x80488dc

08048610 <switch_eg>: . . . 8048622: 77 0c ja 8048630 8048624: ff 24 95 dc 88 04 08 jmp *0x80488dc(,%edx,4)

switch_eg: . . .

ja .L61 # if > goto defaultjmp *.L62(,%edx,4) # goto JTab[x]

Assembly Code

Disassembled Object Code

Page 11: Machine-Level Programming 3 Control Flow

– 11 – CMSC 313, F’09

IA32 Object Code (cont.)Jump Table

Doesn’t show up in disassembled code Can inspect using GDB

gdb asm-cntl

(gdb) x/7xw 0x80488dcExamine 7 hexadecimal format “words” (4-bytes each)Use command “help x” to get format documentation

0x80488dc:

0x08048630

0x08048650

0x0804863a

0x08048642

0x08048630

0x08048649

0x08048649

Page 12: Machine-Level Programming 3 Control Flow

– 12 – CMSC 313, F’09

Disassembled Targets 8048630: bb 02 00 00 00 mov $0x2,%ebx 8048635: 89 d8 mov %ebx,%eax 8048637: 5b pop %ebx 8048638: c9 leave 8048639: c3 ret 804863a: 8b 45 0c mov 0xc(%ebp),%eax 804863d: 99 cltd 804863e: f7 f9 idiv %ecx 8048640: 89 c3 mov %eax,%ebx 8048642: 01 cb add %ecx,%ebx 8048644: 89 d8 mov %ebx,%eax 8048646: 5b pop %ebx 8048647: c9 leave 8048648: c3 ret 8048649: 29 cb sub %ecx,%ebx 804864b: 89 d8 mov %ebx,%eax 804864d: 5b pop %ebx 804864e: c9 leave 804864f: c3 ret 8048650: 8b 5d 0c mov 0xc(%ebp),%ebx 8048653: 0f af d9 imul %ecx,%ebx 8048656: 89 d8 mov %ebx,%eax 8048658: 5b pop %ebx 8048659: c9 leave 804865a: c3 ret

Page 13: Machine-Level Programming 3 Control Flow

– 13 – CMSC 313, F’09

Matching Disassembled Targets 8048630: bb 02 00 00 00 mov 8048635: 89 d8 mov 8048637: 5b pop 8048638: c9 leave 8048639: c3 ret 804863a: 8b 45 0c mov 804863d: 99 cltd 804863e: f7 f9 idiv 8048640: 89 c3 mov 8048642: 01 cb add 8048644: 89 d8 mov 8048646: 5b pop 8048647: c9 leave 8048648: c3 ret 8048649: 29 cb sub 804864b: 89 d8 mov 804864d: 5b pop 804864e: c9 leave 804864f: c3 ret 8048650: 8b 5d 0c mov 8048653: 0f af d9 imul 8048656: 89 d8 mov 8048658: 5b pop 8048659: c9 leave 804865a: c3 ret

0x08048630

0x08048650

0x0804863a

0x08048642

0x08048630

0x08048649

0x08048649

Page 14: Machine-Level Programming 3 Control Flow

– 14 – CMSC 313, F’09

Sparse Switch Example

Not practical to use jump tableWould require 1000

entries Obvious translation into

if-then-else would have max. of 9 tests

/* Return x/111 if x is multiple && <= 999. -1 otherwise */int div111(int x){ switch(x) { case 0: return 0; case 111: return 1; case 222: return 2; case 333: return 3; case 444: return 4; case 555: return 5; case 666: return 6; case 777: return 7; case 888: return 8; case 999: return 9; default: return -1; }}

Page 15: Machine-Level Programming 3 Control Flow

– 15 – CMSC 313, F’09

Sparse Switch Code (IA32)

Compares x to possible case values

Jumps different places depending on outcomes

movl 8(%ebp),%eax # get xcmpl $444,%eax # x:444je .L8jg .L16cmpl $111,%eax # x:111je .L5jg .L17testl %eax,%eax # x:0je .L4jmp .L14

. . .

. . ..L5:

movl $1,%eaxjmp L19

.L6:movl $2,%eaxjmp L19

.L7:movl $3,%eaxjmp L19

.L8:movl $4,%eaxjmp L19. . .

Page 16: Machine-Level Programming 3 Control Flow

– 16 – CMSC 313, F’09

-1

-1 -1-1

Sparse Switch Code Structure

Organizes cases as binary tree Logarithmic performance

444

111 777

0 222 555 888

333 666 9990

1

4

7

5 8

9

2

3 6

<

=

>=

< >=

< >=

< >=

= = =

= =

Page 17: Machine-Level Programming 3 Control Flow

– 17 – CMSC 313, F’09

SummarizingC Control

if-then-else do-while while, for switch

Assembler Control Conditional jump Conditional move Indirect jump

Compiler Must generate assembly

code to implement more complex control

Standard Techniques IA32 loops converted to do-while

form Large switch statements use

jump tables

Conditions in CISC CISC machines generally have

condition code registers

Page 18: Machine-Level Programming 3 Control Flow

– 18 – CMSC 313, F’09

Quiz TimeWrite the C that produced the following assembly language code

This function takes a single integer argument located at 0x8(%ebp)

00000000 <mystery1>:

0: 55 push %ebp

1: 31 c0 xor %eax,%eax

3: 89 e5 mov %esp,%ebp

5: 8b 4d 08 mov 0x8(%ebp),%ecx

8: 85 c9 test %ecx,%ecx

a: 7e 0c jle 18 <mystery1+0x18>

c: 31 d2 xor %edx,%edx

e: 89 f6 mov %esi,%esi

10: 8d 04 50 lea (%eax,%edx,2),%eax

13: 42 inc %edx

14: 39 d1 cmp %edx,%ecx

16: 75 f8 jne 10 <mystery1+0x10>

18: 5d pop %ebp

19: c3 ret

Page 19: Machine-Level Programming 3 Control Flow

– 19 – CMSC 313, F’09

Quiz TimeWrite the C that produced the following assembly language code

This function takes a single integer argument located at 0x8(%ebp)00000000 <mystery2>:

0: 55 push %ebp

1: 31 c0 xor %eax,%eax

3: 89 e5 mov %esp,%ebp

5: 8b 4d 08 mov 0x8(%ebp),%ecx

8: 85 c9 test %ecx,%ecx

a: 7e 0c jle 18 <mystery2+0x18>

c: 31 d2 xor %edx,%edx

e: 89 f6 mov %esi,%esi

10: 8d 04 50 lea (%eax,%edx,2),%eax

13: 42 inc %edx

14: 39 d1 cmp %edx,%ecx

16: 75 f8 jne 10 <mystery2+0x10>

18: 5d pop %ebp

19: c3 ret