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  • When is assembler faster than C?

    - by Adam Bellaire
    One of the stated reasons for knowing assembler is that, on occasion, it can be employed to write code that will be more performant than writing that code in a higher-level language, C in particular. However, I've also heard it stated many times that although that's not entirely false, the cases where assembler can actually be used to generate more performant code are both extremely rare and require expert knowledge of and experience with assembler. This question doesn't even get into the fact that assembler instructions will be machine-specific and non-portable, or any of the other aspects of assembler. There are plenty of good reasons for knowing assembler besides this one, of course, but this is meant to be a specific question soliciting examples and data, not an extended discourse on assembler versus higher-level languages. Can anyone provide some specific examples of cases where assembler will be faster than well-written C code using a modern compiler, and can you support that claim with profiling evidence? I am pretty confident these cases exist, but I really want to know exactly how esoteric these cases are, since it seems to be a point of some contention.

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  • ARMv6 FIQ, acknowledge interrupt

    - by fastmonkeywheels
    I'm working with an i.mx35 armv6 core processor. I have Interrupt 62 configured as a FIQ with my handler installed and being called. My handler at the moment just toggles an output pin so I can test latency with a scope. With the code below, once I trigger the FIQ it continues forever as fast as it can, apparently not being acknowledged. I'm triggering the FIQ by means of the Interrupt Force Register so I'm assured that the source isn't triggering it this fast. If I disable Interrupt 62 in the AVIC in my FIQ routine the interrupt only triggers once. I have read the sections on the VIC Port in the ARM1136JF-S and ARM1136J-S Technical Reference Manual and it covers proper exit procedure. I'm only having one FIQ handler so I have no need to branch. The line that I don't understand is: STR R0, [R8,#AckFinished] I'm not sure what AckFinished is supposed to be or what this command is supposed to do. My FIQ handler is below: ldr r9, IOMUX_ADDR12 ldr r8, [r9] orr r8, #0x08 @ top LED str r8,[r9] @turn on LED bic r8, #0x08 @ top LED str r8,[r9] @turn off LED subs pc, r14, #4 IOMUX_ADDR12: .word 0xFC2A4000 @remapped IOMUX addr My handler returns just fine and normal system operation resumes if I disable it after the first go, otherwise it triggers constantly and the system appears to hang. Do you think my assumption is right that the core isn't acknowledging the AVIC or could there be another cause of this FIQ triggering?

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  • invalid effective address calculation!

    - by Zia ur Rahman
    Hay Dear! Please look at the following program, the error is invalid effective address calculation and i have mentioned that line please tell me why its invalid effective address calculation here is the program [org 0x100] jmp start array1: dw 10,15,20,25,30,35,40,45,50,55 array2: dw 15,10,20,35,40,30,55,50,25,45 start: mov bx,0 mov dx,0 loop: mov ax,[array2+bx] cmp ax,[array1+cx]//here is the error invalid effective address calculation jne NextElementOfArray1 NextElementOfArray2: add bx,2 cmp bx,20 je end mov cx,0 jmp loop NextElementOfArray1: add cx,2 cmp cx,20 je NextElementOfArray2 jmp loop end: mov ax,0x4c00 int 0x21

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  • IA-32: Pushing a byte onto a stack isn't possible on Pentium, why?

    - by Tim Green
    Hi, I've come to learn that you cannot push a byte directly onto the Intel Pentium's stack, can anyone explain this to me please? The reason that I've been given is because the esp register is word-addressable (or, that is the assumption in our model) and it must be an "even address". I would have assumed decrementing the value of some 32-bit binary number wouldn't mess with the alignment of the register, but apparently I don't understand enough. I have tried some NASM tests and come up that if I declare a variable (bite db 123) and push it on to the stack, esp is decremented by 4 (indicating that it pushed 32-bits?). But, "push byte bite" (sorry for my choice of variable names) will result in a kind error: test.asm:10: error: Unsupported non-32-bit ELF relocation Any words of wisdom would be greatly appreciated during this troubled time. I am first year undergraduate so sorry for my naivety in any of this. Tim

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  • MBR Booting from DOS

    - by eflukx
    For a project I would like to invoke the MBR on the first harddisk directly from DOS. I've written a small assembler program that loads the MBR in memory at 0:7c00h an does a far jump to it. I've put my util on a bootable floppy. The disk (HD0, 0x80) i'm trying to boot has a TrueCrypt boot loader on it. It shows up the TrueCrypt screen, but after typing in the password it crashes the system. When I run my little utlility (w00t.com) on a normal WinXP machine it seams to crash immedealty. Apparently I'm forgetting some crucial stuff the BIOS normally does, my guess is it's something trivial. Can someone with better bare-metal DOS and BIOS experience help me out? Heres my code: .MODEL tiny .386 _TEXT SEGMENT USE16 INCLUDE BootDefs.i ORG 100h start: ; http://vxheavens.com/lib/vbw05.html ; Before DOS has booted the BIOS stores the amount of usable lower memory ; in a word located at 0:413h in memory. We going to erase this value because ; we have booted dos before loading the bootsector, and dos is fat (and ugly). ; fake free memory ;push ds ;push 0 ;pop ds ;mov ax, TC_BOOT_LOADER_SEGMENT / 1024 * 16 + TC_BOOT_MEMORY_REQUIRED ;mov word ptr ds:[413h], ax ;ax = memory in K ;pop ds ;lea si, memory_patched_msg ;call print ;mov ax, cs mov ax, 0 mov es, ax ; read first sector to es:7c00h (== cs:7c00) mov dl, 80h mov cl, 1 mov al, 1 mov bx, 7c00h ;load sector to es:bx call read_sectors lea si, mbr_loaded_msg call print lea si, jmp_to_mbr_msg call print ;Set BIOS default values in environment cli mov dl, 80h ;(drive C) xor ax, ax mov ds, ax mov es, ax mov ss, ax mov sp, 0ffffh sti push es push 7c00h retf ;Jump to MBR code at 0:7c00h ; Print string print: xor bx, bx mov ah, 0eh cld @@: lodsb test al, al jz print_end int 10h jmp @B print_end: ret ; Read sectors of the first cylinder read_sectors: mov ch, 0 ; Cylinder mov dh, 0 ; Head ; DL = drive number passed from BIOS mov ah, 2 int 13h jnc read_ok lea si, disk_error_msg call print read_ok: ret memory_patched_msg db 'Memory patched', 13, 10, 7, 0 mbr_loaded_msg db 'MBR loaded', 13, 10, 7, 0 jmp_to_mbr_msg db 'Jumping to MBR code', 13, 10, 7, 0 disk_error_msg db 'Disk error', 13, 10, 7, 0 _TEXT ENDS END start

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  • PE Header Requirements

    - by Pindatjuh
    What are the requirements of a PE file (PE/COFF)? What fields should be set, which value, at a bare minimum for enabling it to "run" on Windows (i.e. executing "ret" instruction and then close, without error). The library I am building first is the linker: Now, the problem I have is the PE file (PE/COFF). I don't know what is "required" for a PE file before it can actually execute on my platform. My testing platform is Vista. I get an error message, saying "This is not a valid Win32 executable." when I execute it by double-clicking, and I get an "Access Denied." when executing it with CLI cmd. I have two sections, .text and .data. I've implemented the PE headers as provided by several online documents, i.e. MSDN and some other thirdparty documentation. If I use a hex-editor, it looks almost like a regular PE file. I don't use any imports, nor IAT, nor any directories in the PE header. Edit: I've added an import table, still not a valid .exe-file, says my Windows. I've tried to use values which are also mentioned at the smallest PE-file guide. No luck. Really the only thing I can't seem to figure out is what is required and what isn't. Some guides tell me everything is required, whilst others say about deprications: and it can be zero. I hope this is enough information. Thank you, in advance. Raw data (as requested) of current PE header: 4D 5A 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 40 00 00 00 50 45 00 00 4C 01 02 00 C8 7A 55 4B 00 00 00 00 00 00 00 00 E0 00 82 01 0B 01 0D 25 00 10 00 00 00 10 00 00 00 00 00 00 00 10 00 00 00 10 00 00 00 20 00 00 00 00 40 00 00 10 00 00 00 02 00 00 01 00 0B 00 00 00 00 00 03 00 0A 00 00 00 00 00 00 22 00 00 38 01 00 00 00 00 00 00 03 00 00 00 00 40 00 00 00 40 00 00 00 40 00 00 00 40 00 00 00 00 00 00 0E 00 00 00 00 00 00 00 00 00 00 00 00 20 00 00 24 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 2E 74 65 78 74 00 00 00 00 00 00 00 00 10 00 00 00 02 00 00 00 02 00 00 00 00 00 00 00 00 00 00 00 00 00 00 20 00 00 60 2E 69 64 61 74 61 00 00 00 00 00 00 00 20 00 00 00 02 00 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 00 40 00 00 C0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 3C 20 00 00 00 00 00 00 00 00 00 00 24 20 00 00 34 20 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 4B 45 52 4E 45 4C 33 32 2E 64 6C 6C 00 00 00 00 01 00 00 80 00 00 00 00 01 00 00 80 00 00 00 00

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  • Combining charcters in mips

    - by Krewie
    Hello , i was wondering if there was anyway of combining two characters to form one character. For instance, i have the character 6 and 7 , i want to combine them and make the result 67 that is saved in a register, is there any solution to this problem ? //Thx in advance

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  • Why do we need different CPU architecture for server & mini/mainframe & mixed-core?

    - by claws
    Hello, I was just wondering what other CPU architectures are available other than INTEL & AMD. So, found List of CPU architectures on Wikipedia. It categorizes notable CPU architectures into following categories. Embedded CPU architectures Microcomputer CPU architectures Workstation/Server CPU architectures Mini/Mainframe CPU architectures Mixed core CPU architectures I was analyzing the purposes and have few doubts. I taking Microcomputer CPU (PC) architecture as reference and comparing others. Embedded CPU architecture: They are a completely new world. Embedded systems are small & do very specific task mostly real time & low power consuming so we do not need so many & such wide registers available in a microcomputer CPU (typical PC). In other words we do need a new small & tiny architecture. Hence new architecture & new instruction RISC. The above point also clarifies why do we need a separate operating system (RTOS). Workstation/Server CPU architectures I don't know what is a workstation. Someone clarify regarding the workstation. As of the server. It is dedicated to run a specific software (server software like httpd, mysql etc.). Even if other processes run we need to give server process priority therefore there is a need for new scheduling scheme and thus we need operating system different than general purpose one. If you have any more points for the need of server OS please mention. But I don't get why do we need a new CPU Architecture. Why cant Microcomputer CPU architecture do the job. Can someone please clarify? Mini/Mainframe CPU architectures Again I don't know what are these & what miniframes or mainframes used for? I just know they are very big and occupy complete floor. But I never read about some real world problems they are trying to solve. If any one working on one of these. Share your knowledge. Can some one clarify its purpose & why is it that microcomputer CPU archicture not suitable for it? Is there a new kind of operating system for this too? Why? Mixed core CPU architectures Never heard of these. If possible please keep your answer in this format: XYZ CPU architectures Purpose of XYZ Need for a new architecture. why can't current microcomputer CPU architecture work? They go upto 3GHZ & have upto 8 cores. Need for a new Operating System Why do we need a new kind of operating system for this kind of archictures?

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  • "Hello World" in less than 20 bytes

    - by xelurg
    We have had an interesting competition once, where everyone would write their implementation of hello world program. One requirement was that is should be less than 20 bytes in compiled form. the winner would be the one whose version is the smallest... What would be your solution? :) Platform: 32bit, x86 OS: DOS, Win, GNU/Linux, *BSD Language: Asm, C, or anything else that compiles into binary executable (i.e. no bash scripts and stuff ;)

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  • Algorithm for finding the smallest power of two that's greater or equal to a given value

    - by Boyan
    I need to find the smallest power of two that's greater or equal to a given value. So far, I have this: int value = 3221; // 3221 is just an example, could be any number int result = 1; while (result < value) result <<= 1; It works fine, but feels kind of naive. Is there a better algorithm for that problem? EDIT. There were some nice Assembler suggestions, so I'm adding those tags to the question.

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  • Subroutine & GoTo design

    - by sub
    I have a strange question concerning subroutines: As I'm creating a minimal language and I don't want to add high-level loops like while or for I was planning on just adding gotos to keep it Turing-Complete. Now I thought, eww - gotos - I wouldn't want to program in that language if I had to use gotos so often. So I thought about adding subroutines instead. I see the difference as the following: gotos Go to (captain obvious) a previously defined point and continue executing the program from there. Leads to hardly understandable and buggy code, I think that's a fact. subroutines Similiar: You define their starting point somewhere, as you call them the program jumps there - but the subroutine can go back to the point it was called from with return. Okay. Why didn't I just add the more function-like, nice looking subroutines? Because: In order to make return work if I call subroutines from within subroutines from within other subroutines, I'd have to use a stack containing the point where the currently running subroutine came from at top. That would then mean that I would, if I create loops using the subroutines, end up with an extremely memory-eating, overflowing stack with return locations. Not good. Don't think of my subroutines as functions. They are just gotos that return to the point they were called from, they don't actually give back values like the return x; statement in nearly all today's languages. Now to my actual questions: How can I solve the above problem with the stack overflow on loops with subroutines? Do I have to add a separate goto language construct without the return option? Assembler doesn't have loops but as I have seen myJumpPoint:, jnz, jz, retn. That means to me that there must also be a stack containing all the return locations. Am I right with that? What about long running loops then? Don't they overflow the stack/eat memory then? Am I getting the retn symbol in assembler totally wrong? If yes, please explain it to me.

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  • Working with ieee format numbers in ARM

    - by Jake Sellers
    I'm trying to write an ARM program that will convert an ieee number to a TNS format number. TNS is a format used by some super computers, and is similar to ieee but different. I'm trying to use several masks to place the three different "part" of the ieee number in separate registers so I can move them around accordingly. Here is my unpack subroutine: UnpackIEEE LDR r1, SMASK ;load the sign bit mask into r1 LDR r2, EMASK ;load the exponent mask into r2 LDR r3, GMASK ;load the significand mask into r3 AND r4, r0, r1 ;apply sign mask to IEEE and save into r4 AND r5, r0, r2 ;apply exponent mask to IEEE and save into r5 AND r6, r0, r3 ;apply significand mask to IEEE and save into r6 MOV pc, r14 ;return And here are the masks and number declarations so you can understand: IEEE DCD 0x40300000 ;2.75 decimal or 01000000001100000000000000000000 binary SMASK DCD 0x80000000 ;Sign bit mask EMASK DCD 0x7F800000 ;Exponent mask GMASK DCD 0x007FFFFF ;Significand mask When I step through with the debugger, the results I get are not what I expect after working through it on paper. EDIT: What I mean, is that after the subroutine runs, registers 4, 5, and 6 all remain 0. I can't figure out why the masks are not working. I think I do not fully understand how the number is being stored in the register or using the masks wrong. Any help appreciated. If you need more info just ask. EDIT: entry point: Very simple, just trying to get these subroutines working. ENTRY LDR r1, IEEE ;load IEEE num into r1 BL UnpackIEEE ;call unpack sub SWI SWI_Exit ;finish

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  • [ebp + 6] instead of +8 in a JIT compiler

    - by David Titarenco
    I'm implementing a simplistic JIT compiler in a VM I'm writing for fun (mostly to learn more about language design) and I'm getting some weird behavior, maybe someone can tell me why. First I define a JIT "prototype" both for C and C++: #ifdef __cplusplus typedef void* (*_JIT_METHOD) (...); #else typedef (*_JIT_METHOD) (); #endif I have a compile() function that will compile stuff into ASM and stick it somewhere in memory: void* compile (void* something) { // grab some memory unsigned char* buffer = (unsigned char*) malloc (1024); // xor eax, eax // inc eax // inc eax // inc eax // ret -> eax should be 3 /* WORKS! buffer[0] = 0x67; buffer[1] = 0x31; buffer[2] = 0xC0; buffer[3] = 0x67; buffer[4] = 0x40; buffer[5] = 0x67; buffer[6] = 0x40; buffer[7] = 0x67; buffer[8] = 0x40; buffer[9] = 0xC3; */ // xor eax, eax // mov eax, 9 // ret 4 -> eax should be 9 /* WORKS! buffer[0] = 0x67; buffer[1] = 0x31; buffer[2] = 0xC0; buffer[3] = 0x67; buffer[4] = 0xB8; buffer[5] = 0x09; buffer[6] = 0x00; buffer[7] = 0x00; buffer[8] = 0x00; buffer[9] = 0xC3; */ // push ebp // mov ebp, esp // mov eax, [ebp + 6] ; wtf? shouldn't this be [ebp + 8]!? // mov esp, ebp // pop ebp // ret -> eax should be the first value sent to the function /* WORKS! */ buffer[0] = 0x66; buffer[1] = 0x55; buffer[2] = 0x66; buffer[3] = 0x89; buffer[4] = 0xE5; buffer[5] = 0x66; buffer[6] = 0x66; buffer[7] = 0x8B; buffer[8] = 0x45; buffer[9] = 0x06; buffer[10] = 0x66; buffer[11] = 0x89; buffer[12] = 0xEC; buffer[13] = 0x66; buffer[14] = 0x5D; buffer[15] = 0xC3; // mov eax, 5 // add eax, ecx // ret -> eax should be 50 /* WORKS! buffer[0] = 0x67; buffer[1] = 0xB8; buffer[2] = 0x05; buffer[3] = 0x00; buffer[4] = 0x00; buffer[5] = 0x00; buffer[6] = 0x66; buffer[7] = 0x01; buffer[8] = 0xC8; buffer[9] = 0xC3; */ return buffer; } And finally I have the main chunk of the program: void main (int argc, char **args) { DWORD oldProtect = (DWORD) NULL; int i = 667, j = 1, k = 5, l = 0; // generate some arbitrary function _JIT_METHOD someFunc = (_JIT_METHOD) compile(NULL); // windows only #if defined _WIN64 || defined _WIN32 // set memory permissions and flush CPU code cache VirtualProtect(someFunc,1024,PAGE_EXECUTE_READWRITE, &oldProtect); FlushInstructionCache(GetCurrentProcess(), someFunc, 1024); #endif // this asm just for some debugging/testing purposes __asm mov ecx, i // run compiled function (from wherever *someFunc is pointing to) l = (int)someFunc(i, k); // did it work? printf("result: %d", l); free (someFunc); _getch(); } As you can see, the compile() function has a couple of tests I ran to make sure I get expected results, and pretty much everything works but I have a question... On most tutorials or documentation resources, to get the first value of a function passed (in the case of ints) you do [ebp+8], the second [ebp+12] and so forth. For some reason, I have to do [ebp+6] then [ebp+10] and so forth. Could anyone tell me why?

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  • Setting a cell's format using Excel 2007 Interop and C#

    - by CVertex
    I'm using the office 2007 interop assemblies to create some excel spreadsheets. There are plenty of questions on here about getting started and MSDN contains heaps of articles, like this one. The API is funky, and sometimes a bit confusing. When I set a value of a cell, is there a way to set it's format? I'd like to mark particular fields as Date's so my customer can run excel macros on them. Also, numbers would be useful. Thanks!

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  • FASM vc MASM trasnlation problem in mov si, offset msg

    - by Ruben Trancoso
    hi folks, just did my first test with MASM and FASM with the same code (almos) and I falled in trouble. The only difference is that to produce just the 104 bytes I need to write to MBR in FASM I put org 7c00h and in MASM 0h. The problem is on the mov si, offset msg that in the first case transletes it to 44 7C (7c44h) and with masm translates to 44 00 (0044h)! but just when I change org 7c00h to org 0h in MASM. Otherwise it will produce the entire segment from 0 to 7dff. how do I solve it? or in short, how to make MASM produce a binary that begins at 7c00h as it first byte and subsequent jumps remain relative to 7c00h? .model TINY .code org 7c00h ; Boot entry point. Address 07c0:0000 on the computer memory xor ax, ax ; Zero out ax mov ds, ax ; Set data segment to base of RAM jmp start ; Jump to the first byte after DOS boot record data ; ---------------------------------------------------------------------- ; DOS boot record data ; ---------------------------------------------------------------------- brINT13Flag db 90h ; 0002h - 0EH for INT13 AH=42 READ brOEM db 'MSDOS5.0' ; 0003h - OEM name & DOS version (8 chars) brBPS dw 512 ; 000Bh - Bytes/sector brSPC db 1 ; 000Dh - Sectors/cluster brResCount dw 1 ; 000Eh - Reserved (boot) sectors brFATs db 2 ; 0010h - FAT copies brRootEntries dw 0E0h ; 0011h - Root directory entries brSectorCount dw 2880 ; 0013h - Sectors in volume, < 32MB brMedia db 240 ; 0015h - Media descriptor brSPF dw 9 ; 0016h - Sectors per FAT brSPH dw 18 ; 0018h - Sectors per track brHPC dw 2 ; 001Ah - Number of Heads brHidden dd 0 ; 001Ch - Hidden sectors brSectors dd 0 ; 0020h - Total number of sectors db 0 ; 0024h - Physical drive no. db 0 ; 0025h - Reserved (FAT32) db 29h ; 0026h - Extended boot record sig brSerialNum dd 404418EAh ; 0027h - Volume serial number (random) brLabel db 'OSAdventure' ; 002Bh - Volume label (11 chars) brFSID db 'FAT12 ' ; 0036h - File System ID (8 chars) ;------------------------------------------------------------------------ ; Boot code ; ---------------------------------------------------------------------- start: mov si, offset msg call showmsg hang: jmp hang msg db 'Loading...',0 showmsg: lodsb cmp al, 0 jz showmsgd push si mov bx, 0007 mov ah, 0eh int 10h pop si jmp showmsg showmsgd: retn ; ---------------------------------------------------------------------- ; Boot record signature ; ---------------------------------------------------------------------- dw 0AA55h ; Boot record signature END

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  • setcontext and makecontext to call a generic function pointer

    - by Simone Margaritelli
    In another question i had the problem to port the code unsigned long stack[] = { 1, 23, 33, 43 }; /* save all the registers and the stack pointer */ unsigned long esp; asm __volatile__ ( "pusha" ); asm __volatile__ ( "mov %%esp, %0" :"=m" (esp)); for( i = 0; i < sizeof(stack); i++ ){ unsigned long val = stack[i]; asm __volatile__ ( "push %0" :: "m"(val) ); } unsigned long ret = function_pointer(); /* restore registers and stack pointer */ asm __volatile__ ( "mov %0, %%esp" :: "m" (esp) ); asm __volatile__ ( "popa" ); To a 64bit platform and many guys told me i should use the setcontext and makecontext functions set instead due to the calling conversion differences between 32 and 64 bits and portability issues. Well, i really can't find any useful documentation online, or at least not the kind i need to implement this, so, how can i use those functions to push arguments onto the stack, call a generic function pointer, obtain the return value and then restore the registers?

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  • Implementing a Mutex Lock in C

    - by Adam
    I'm trying to make a really mutex in C and for some reason I'm getting cases where two threads are getting the lock at the same time, which shouldn't be possible. Any ideas why it's not working? void mutexLock(mutex_t *mutexlock, pid_t owner) { int failure; while(mutexlock->mx_state == 0 || failure || mutexlock->mx_owner != owner) { failure = 1; if (mutexlock->mx_state == 0) { asm( "test:" "movl $0x01,%%eax\n\t" // move 1 to eax "xchg %%eax,%0\n\t" // try to set the lock bit "mov %%eax,%1\n\t" // export our result to a test var "test %%eax,%%eax\n\t" "jnz test\n\t" :"=r"(mutexlock->mx_state),"=r"(failure) :"r"(mutexlock->mx_state) :"%eax" ); } if (failure == 0) { mutexlock->mx_owner = owner; //test to see if we got the lock bit } } }

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  • how do addressing modes work on a physical level?

    - by altvali
    I'm trying to learn this basic thing about processors that should be taught in every CS department of every university. Yet i can't find it on the net (Google doesn't help) and i can't find it in my class materials either. Do you know any good resource on how addressing modes work on a physical level? I'm particularly interested in Intel processors.

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