A few errors to deal with
in [ASM]
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From: Mint on 26 May 2010 06:06 I am having some problems with this code. Can someone help? Thanks. Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland International Assembling file: timeit.asm **Error** c:\16_bit\monitor.asm(69) Code or data emission to undeclared segment **Error** c:\16_bit\monitor.asm(85) Name must come first **Error** c:\16_bit\monitor.asm(89) Illegal instruction **Error** c:\16_bit\monitor.asm(90) Operand types do not match **Error** c:\16_bit\monitor.asm(91) Operand types do not match **Error** c:\16_bit\monitor.asm(92) Can't use this outside macro **Error** c:\16_bit\monitor.asm(109) Name must come first **Error** c:\16_bit\monitor.asm(113) Illegal instruction **Error** c:\16_bit\monitor.asm(114) Operand types do not match **Error** c:\16_bit\monitor.asm(115) Operand types do not match **Error** c:\16_bit\monitor.asm(122) Can't use this outside macro **Error** timeit.asm(21) Undefined symbol: MONITOR_INIT **Error** timeit.asm(22) Illegal instruction **Error** timeit.asm(24) Illegal instruction ..model small ..586 ..stack 200h include c:\16_bit\monitor.asm ..data time_cycles dq ? cycle dd ? cpuid_cycle dd ? mycountlow dd ? mycounthigh dd ? ..code start: mov ax,@data mov ds,ax call monitor_init time_start xor ax,ax time_stop mov [mycountlow],eax mov [mycounthigh],edx exit: mov ax,4c00h int 21h end start ------------------------------------------ ; monitor.asm ; Performance monitoring package ; ; define PProPII if your CPU is a Pentium Pro or a Pentium II ; ; implements: ; ; monitor_init ; initializes the package ; ; time_start ; start cycle count here ; ; time_stop ; stop counting here ; ; note: ; the package can not do nested measurements, since the macro ; returns all cycles in the same variable ; ; ; define cpuid and rdtsc instructions via macros ; this is not necessary is your assembler supports them ; ; MACRO cpuid ; db 0fh,0a2h ; ENDM ; ; MACRO rdtsc ; db 0fh,031h ; ENDM ; ; monitor_init: ; ; input: ; nothing ; ; output: ; cpuid_cycle = initialized to exection time of cpuid ; ; destroys: ; nothing ; monitor_init: IFDEF PProPII pushfd pushad mov ecx,3 getcpuidtime: cpuid rdtsc mov [cycle],eax cpuid rdtsc sub eax,[cycle] mov [cpuid_cycle],eax dec ecx jnz getcpuidtime popad popf ENDIF ret ; ; time_start - start timing point here ; ; input: ; none ; ; output: ; time_cycles initialized ; ; destroys: ; eax, ebx, ecx, edx ; eflags ; MACRO time_start IFDEF PProPII cpuid ENDIF rdtsc mov [time_cycles],eax mov [time_cycles+4],edx ENDM ; ; time_stop - stop timing point here ; ; input: ; none ; ; output: ; eax = low cycle count ; edx = high cycle count ; ; destroys: ; eax, ebx, ecx, edx ; eflags ; MACRO time_stop IFDEF PProPII cpuid ENDIF rdtsc sub eax,[time_cycles] sbb edx,[time_cycles+4] IFDEF ProPII sub eax,[cpuid_cycle] sbb edx,0 ENDIF ENDM
From: io_x on 26 May 2010 13:58 "Mint" <chocolatemint77581(a)yahoo.com> ha scritto nel messaggio news:e3de4069-c526-4c78-9fc5-599185d4f71b(a)s41g2000vba.googlegroups.com... >I am having some problems with this code. > Can someone help? > > Thanks. > > Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland > International > > Assembling file: timeit.asm > **Error** c:\16_bit\monitor.asm(69) Code or data emission to > undeclared segment > **Error** c:\16_bit\monitor.asm(85) Name must come first > **Error** c:\16_bit\monitor.asm(89) Illegal instruction > IFDEF PProPII > pushfd > pushad > > mov ecx,3 > getcpuidtime: > cpuid > rdtsc > mov [cycle],eax > cpuid > rdtsc > sub eax,[cycle] > mov [cpuid_cycle],eax > dec ecx > jnz getcpuidtime > > popad > popf i think above should be popfd but i not understand much so could be not true (if popfd==popf==32bit instruction pop all the 32bit falg register) but the flag register is 32Bits for one 386 cpu? > ENDIF > ret >
From: Frank Kotler on 26 May 2010 16:24 Mint wrote: > I am having some problems with this code. > Can someone help? > > Thanks. > > Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland > International FFS, Andy, get yourself an assembler! .... > time_cycles dq ? ..... > mov [time_cycles],eax > mov [time_cycles+4],edx dword ptr > sub eax,[time_cycles] > sbb edx,[time_cycles+4] dword ptr http://www.japheth.de/JWasm.html Are you *new* at this??? Best, Frank
From: Mint on 27 May 2010 10:57 On May 26, 3:24 pm, Frank Kotler <fbkot...(a)myfairpoint.net> wrote: > Mint wrote: > > I am having some problems with this code. > > Can someone help? > > > Thanks. > > > Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland > > International > > FFS, Andy, get yourself an assembler! > Are you *new* at this??? > > Best, > Frank Be nice. :-) Nothing wrong with Tasm. ; ?? Showing 340 cycles for any block of code ; timeit.asm TASM code ; Help from Shoorick, ; time = cycles/frequency ..model small ..486 ..stack 200h ..data time_cycles dd ? cycle dd ? cpuid_cycle dd ? mycountlow dd ? mycounthigh dd ? ..code include monitor.asm start: mov ax,@data mov ds,ax call monitor_init time_start mov ax,45000 ; ax:dx contains 90,000,000 ; ax = 4a80h ; dx = 55dh ; so, 55d4a80h = 90,000,000 mov bx,2000 mul bx time_stop mov [mycountlow],eax mov [mycounthigh],edx ;mov eax,123456789 ; print low cycle count call PrtDec xor eax,eax ; clear eax ; print high cycle count mov eax,edx call PrtDec mov ax,4c00h int 21h ; Print any number in eax PrtDec proc push eax push ecx push edx mov ecx,0ffffffffh push ecx mov ecx,10 pd1: mov edx,0 div ecx add dl,30h push edx cmp eax,0 jne pd1 pd2: pop edx cmp edx,0ffffffffh je pd3 mov ah,2 int 21h jmp pd2 pd3: pop edx pop ecx pop eax ret PrtDec endp ;exit: ;mov ax,4c00h ;int 21h end start
From: Frank Kotler on 27 May 2010 14:11
Mint wrote: > On May 26, 3:24 pm, Frank Kotler <fbkot...(a)myfairpoint.net> wrote: >> Mint wrote: >>> I am having some problems with this code. >>> Can someone help? >>> Thanks. >>> Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland >>> International >> FFS, Andy, get yourself an assembler! > >> Are you *new* at this??? >> >> Best, >> Frank > > Be nice. :-) > Nothing wrong with Tasm. You're right. I'm sorry. > ; ?? Showing 340 cycles for any block of code Thought you said there was nothing wrong with it. :) > ; timeit.asm TASM code > ; Help from Shoorick, > ; time = cycles/frequency > .model small > .486 > .stack 200h > > .data > > time_cycles dd ? Unless you've changed the hidden-away part, you're still treating this as a qword. This is "wrong", but you're just overwriting "cycle" with the high dword, and I don't think you use "cycle" after "monitor_init" (if then)... so I don't think it matters... > cycle dd ? > cpuid_cycle dd ? > mycountlow dd ? > mycounthigh dd ? > > .code > > include monitor.asm > > start: > > mov ax,@data > mov ds,ax > > call monitor_init What does this do? As I read it, if you haven't got "PProPII" defined, it doesn't do squat. In fact, you don't appear to do "cpuid" at all, if this isn't defined. You probably do want "cpuid", if your machine supports it - using Tasm, ya never know. :) I sympathize with your problems. My attempts to time anything have been "inconclusive". If you get this thing working, time "das" for me! I've got some code... very similar to what you've got here... which "works", I guess, but gives some weird results. "das" is all over the place. "push eax"/"pop eax" executes faster if you do it twice than if you do it once. Things like that make me wonder if there's a bug in the code that I haven't found, or if my PIV is just weird. I'd post it, but it is specific to Linux... Hell, I'll post it anyway. I'll try to work up a more "portable" version, don't waste too much "time" on this... Best, Frank ; uses Herbert's "direct to elf executable" code ; nasm -f bin myprog.asm > myprog.map ; chmod +x myprog ; "tune" this so an empty loop gives zero, most often, ; on your machine. THIS_EMPTY_LOOP equ 384 ; define this on the command line to Nasm, ; or uncomment it here, if you prefer... ; REPT_COUNT equ 1 ; 10000 ;[map all] ;=========================================================================== bits 32 ORIGIN equ 8048000h org ORIGIN section .text code_offset equ 0 code_addr: ;--------------------------- ELF header ----------------------------------- dd $464c457f,$00010101,0,0,$00030002,1,main,$34,0,0,$00200034,2,0 dd 1,code_offset,code_addr,code_addr,code_filez,code_memsz,5,4096 dd 1,data_offset,data_addr,data_addr,data_filez,data_memsz,6,4096 main: ;--------- your code goes here ------------------------------------------ xor eax, eax cpuid rdtsc push edx push eax ;------------------ ; code to time ;times REPT_COUNT mov eax, 0 ;times REPT_COUNT xor eax, eax ;times REPT_COUNT dec eax ; times REPT_COUNT sub eax, byte 1 %rep REPT_COUNT ;push eax ;pop eax ;mov eax, [esp] ;mov [esp], eax ; das %endrep ;------------------- xor eax, eax cpuid rdtsc pop ebx sub eax, ebx pop ecx sbb edx, ecx sub eax, THIS_EMPTY_LOOP sbb edx, 0 mov edi, ascbuf call u64toda ; will need a suitable "printstring" and "exit" for 'doze mov al, 10 ; terminate buffer with LF stosb mov ecx, ascbuf mov edx, edi sub edx, ecx mov ebx, 1 mov eax, 4 int 80h exit: mov eax, 1 int 80h ;---------------- ;-------------------------------------------------------------- ; u64toda - converts (64 bit) integer in edx:eax ; to (comma delimited) decimal representation in ; string in buffer pointed to by edi ; returns edi -> next position in buffer ;----------------------------------------------------------------- u64toda: push eax push ebx push ecx push edx push esi mov ebx, edx ; stash high dword mov esi, 0Ah ; prepare to divide by 10 xor ecx, ecx ; zero the digit count jmp highleft ; check is high word 0 ? highword: xchg eax, ebx ; swap high & low words xor edx, edx ; zero edx for the divide! div esi ; divide high word by 10 xchg eax, ebx ; swap 'em back div esi ; divide low word including remainder push edx ; remainder is our digit - save it inc ecx ; count digits highleft: test ebx, ebx jnz highword lowleft: xor edx, edx ; zero high word div esi ; divide low word by 10 push edx ; our digit inc ecx ; count it or eax, eax ; 0 yet ? jne lowleft cmp ecx, 4 ; commas needed ? jl write2buf ; nope xor edx, edx ; zero high word for divide mov eax, ecx ; number of digits mov ebx, 3 div ebx mov esi, edx ; remainder = number digits before comma test edx, edx jnz write2buf ; no remainder? mov esi, 3 ; we can write 3 digits, then. write2buf: pop eax ; get digit back - in right order add al, '0' ; convert to ascii character stosb ; write it to our buffer dec esi ; digits before comma needed jnz moredigits ; no comma needed yet cmp ecx, 2 ; we at the end? jl moredigits ; don't need comma mov al, ',' ; write a comma stosb mov esi, 3 ; we're good for another 3 digits moredigits: loop write2buf ; write more digits - cx of 'em pop esi pop edx pop ecx pop ebx pop eax ret ;------------------------------------------------------------- ;------------ constant data --------------------------------- ; (note that we're in .text, not .rdata) align 4 ; we have none ;--------------------------------------------------------------------------- align 4 code_memsz equ $ - $$ code_filez equ code_memsz data_addr equ (ORIGIN+code_memsz+4095)/4096*4096 + (code_filez % 4096) data_offset equ code_filez section .data vstart=data_addr ;------------ initialized data ------------------------------ ; we have none ;--------------------------------------------------------------------------- idat_memsz equ $ - $$ bss_addr equ data_addr + ($ - $$) section .bss vstart=bss_addr ;--------------------------- uninitialized data ---------------------------- ascbuf resb 20h ;--------------------------------------------------------------------------- udat_memsz equ $ - $$ data_memsz equ idat_memsz + udat_memsz data_filez equ idat_memsz ;=========================================================================== |