Lecture 3: control transfer instructions

Postacademic course in ICT Lecture 3: control transfer instructions ICT Software & Data processing Part I: The hardware-software interface Module 1: Computer architecture Prof. Koen De Bosschere Electronics Dept Ghent University

Overview • Jumps • Loops • Procedure call and return • Interrupts • System operations • Measuring performance • Instruction decoding • Compilers, linkers, and loaders

Jumps • Unconditional jumps • Conditional jumps • Computed jumps

Unconditional Jumps 10 i1 jmp 24 i3 i4 i5 i6 i7 i8 jmp 20 i10 14 jmp address 18 1c 20 24 28 2c 30 34

Conditional jumps i1 jle 24 10 i1 jle 24 i3 i4 jmp 2c i6 i7 i8 i9 i10 14 18 1c i3 i4 jmp 2c i6 i7 20 24 28 2c i8 i9 i10 30 Basic blocks 34

Jump conditions (1) instruction jump jz jump if zero jc jump if carry jo jump if overflow js jump if sign jnz jump if not zero jnc jump if not carry jno jump if not overflow jns jump if not sign

Jump conditions (2) instruction jump jg jnle jump if greater jge jnl jump if greater or equal jl jnge jump if less jle jng jump if less or equal je jump if equal ja jnbe jump if above jae jnb jump if above or equal jb jnae jump if below jbe jna jump if below or equal 2’s complement binary

Static vs. Computed address jmp 100 mov ebx,100 jmp ebx

Absolute vs. relative address 10 i1 jmp 24 i3 i4 i5 i6 i7 i8 jmp pc-10 i10 14 • Absolute jump to address n • Relative jump n bytes further/back 18 1c 20 24 28 2c 30 34

2c Position independent code 10 10 14 i1 jmp 24 i3 i4 i5 i6 i7 i8 jmp pc-10 i10 14 18 i1 jmp 24 i3 i4 i5 i6 i7 i8 jmp pc-10 i10 18 1c 1c 20 20 24 24 28 28 2c 2c 30 30 34 34 38 3c

loop Loop instruction 10 loop address i1 i2 mov ecx,5 i4 i5 i6 i7 i8 loop 1c i10 14 Decrement ecx Jump to address if ecx <> 0 18 1c 20 24 28 2c 30 34

loop Programmed loop 10 i1 i2 mov ecx,5 i4 i5 i6 i7 i8 sub ecx,1 jnz 1c 14 18 1c 20 24 28 2c 30 34

Function call 10 i1 i1 i2 call 34 14 i2 18 call 34 i7 i8 i9 i10 ret 22 i4 26 i5 30 i6 34 i7 38 i4 i5 i6 i8 42 i9 46 i10 50 ret ca5-15

Function call and return int fivefold(int n) { if (n > 0) return n * 5; else return 0; } int g; main() { g = fivefold(6); }

s=? z=? Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax ???????? 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx ???????? edx ???????? esp 00000108 eip 0000030f ???????? 100 ???????? 104 ???????? 108

s=? z=? Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 00000006 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx ???????? edx ???????? esp 00000108 eip 00000314 push 319 jmp 300 ???????? 100 ???????? 104 ???????? 108

s=? z=? Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 00000006 300: 302: 304: 306: 307: 30c: 30e: 30f: 314` 319: ebx ???????? edx ???????? esp 00000104 eip 00000300 ???????? 100 00000319 104 ???????? 108

s=0 z=0 Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 00000006 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx ???????? edx ???????? esp 00000104 eip 00000302 ???????? 100 00000319 104 ???????? 108

s=0 z=0 Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 00000006 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx ???????? edx ???????? esp 00000104 eip 00000307 ???????? 100 00000319 104 ???????? 108

s=0 z=0 Code fivefold: cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 00000006 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx 00000005 edx ???????? esp 00000104 eip 0000030c ???????? 100 00000319 104 ???????? 108

s=0 z=0 Code fivefold : cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 0000001e 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx 00000005 edx 00000000 esp 00000104 eip 0000030e ???????? 100 00000319 104 ???????? 108

s=0 z=0 Code fivefold : cmp eax,0 jg positive xor eax, eax ret positive: mov ebx, 5 imul ebx ret main: mov eax, 6 call fivefold mov g, eax eax 0000001e 300: 302: 304: 306: 307: 30c: 30e: 30f: 314: 319: ebx 00000005 edx 00000000 esp 00000108 eip 00000319 ???????? 100 00000319 104 ???????? 108

s=0 z=0 Saving registers fivefold : push edx cmp eax,0 jg positive xor eax, eax pop edx ret positive: mov ebx, 5 imul ebx pop edx ret main: mov eax, 6 call fivefold mov g, eax eax 0000001e 300: 301: 303: 305: 307: 308 309: 30e: 310: 311: 312: 317: 31c: ebx 00000005 edx ???????? esp 00000108 ???????? 0fc edx 100 0000031c 104 ???????? 108

fivefold: push edx push ebx cmp eax,0 jg positive xor eax, eax pop ebx pop edx ret positive: mov ebx, 5 imul ebx pop ebx pop edx ret main: mov eax, 6 call fivefold mov g, eax s=0 z=0 Saving registers eax 0000001e ebx ???????? edx ???????? ???????? 0f8 ebx 0fc edx 100 ret address 104 ???????? 108

Control Flow Graph fivefold: push edx push ebx cmp eax,0 jg positive xor eax, eax pop ebx pop edx ret positive: mov ebx, 5 imul ebx pop ebx pop edx ret

Control Flow Graph fivefold: cmp eax,0 jg positive xor eax, eax ret positive: push edx push ebx mov ebx, 5 imul ebx pop ebx pop edx ret

Control Flow Graph fivefold: cmp eax,0 jg positive xor eax, eax ret positive: push edx mov edx, 5 imul edx pop edx ret

Parameter Passing via stack fivefold: mov eax,[esp+4] cmp eax,0 jg positive xor eax, eax ret 4 positive: mov ebx, 5 imul ebx ret 4 main: push 6 call fivefold mov g, eax ???????? 0fc esp ret address 100 00000006 104 ???????? 108

Local Variables int fivefold(int n) { int result; if (n > 0) result = n * 5; else result = 0; return result; } int g; main() { g = fivefold(6); }

fivefold: sub esp,4 cmp eax,0 jg positive xor eax,eax mov [esp], eax jmp end positive: mov ebx, 5 imul ebx mov [esp], eax end: mov eax,[esp] add esp,4 ret main: mov eax, 6 call fivefold mov g, eax Local Variables ???????? 0fc esp ???????? 100 ret address 104 ???????? 108

Control Flow Graph fivefold: sub esp,4 cmp eax,0 jg positive positive: mov ebx, 5 imul ebx mov [esp],eax xor eax, eax mov [esp],eax jmp end end: mov eax,[esp] add esp,4 ret

fivefold: push ebx push edx sub esp, 4 mov eax, [esp+10] cmp eax,0 jg positive xor eax,eax mov [esp], eax jmp end positive: mov ebx, 5 imul ebx mov [esp], eax end: mov eax,[esp] add esp, 4 pop edx pop ebx ret 4 main: push 6 call fivefold mov g, eax Complete Picture stack frame esp result esp+4 edx esp+8 ebx esp+c ret address esp+10 00000006

Stack frames p1 p2 p3 p4 Activation tree

Interrupts • Jump to a routine via a number instead of an address • The addresses of the routines are stored in an address table (the so-called vector table) • Used to catch errors, or as an interface to the operating system

i1 i2 i3 i4 int 3 i5 i1 i6 i2 i7 i3 i8 i4 int 2 i5 i10 i6 i11 Interrupt routine 3 t Interrupts Vector table 0 1 2 3 4

System operations • Controlling the machine: manipulation of the processor state • interrupts on/off • changing the privilege level • halt instruction • switching from big-endian to little-endian • memory management (caches, virtual memory, etc.)

MIPS & MFLOPS • MIPS: Million instructions per second • MFLOPS: Million floating point operations per second • Problems: • Depends on the architecture (multiply accumulate = 1 or 2 instructions?) • Depends on de the program

Benchmark programs • The only reliable performance metrics = execution time • Always mention the program + input? • Ideally: your own application • Difficult to realize (porting) • That’s why ‘typical programs’ are used: benchmarks

Benchmark programs • Whetstone, Dhrystone • EEMBC, Mediabench • TPC-benchmarks • SPEC (Standard Performance Evaluation Corporation): Spec92, Spec95, Spec2000, Spec2006 (SpecInt, SpecFp, SpecRate) • Ziff-Davis

1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 data 2 w word opcode add immediate to accumulator Instruction decoding add ax,2 05 02 00 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0

ea=514 sub r,r sub r,m sub m,r 8088 3 13+ea 24+ea 8086 3 9+ea 16+ea 80286 2 7 7 80386 2 7 6 80486 1 2 3 Pentium 1 2 3 Instruction description SUB subtract O D I T S Z A P C 001010dwoorrrmmm disp * * * * * *

Program development source object file compiler object file object files executable file linker libraries

Lecture 3: control transfer instructions