It looks like you are trying to do something similar to this: #include <stdio.h> double hullSpeed(double lgth) { double result; __asm__( “fldl %1\n\t” //st(0)=>st(1), st(0)=lgth . FLDL means load double float “fsqrt\n\t” //st(0) = square root st(0) “fmulp\n\t” //Multiplies st(0) and st(1) (1.34). Result in st(0) : “=&t” (result) : “m” (lgth), “0” (1.34)); return … Read more
Assuming Intel syntax, where MOV AX, EBX if allowed, would copy the contents of EBX to AX. Just try it. As I remember it’s not supported. But you can copy any 16-bit group, e.g. MOV AX, BX However, regarding the opposite, extending a bit pattern, like the hypothetical MOV EBX, AX how to do that … Read more
First of all, C functions are call-by-value: the int x arg in the function is a copy. Modifying it doesn’t modify the caller’s copy of whatever they passed, so your swap makes zero sense. Second, you’re using the return value of the function, but you don’t have a return statement. In C (unlike C++), it’s … Read more
Is it possibile, given a sequence of bytes x86 instructions into a stream of random bytes, to decode their instructions? Yes. Many kinds of processors do it. It is one of the easiest task they have. Are opcodes of a fixed length No. or is there any way to detect those instructions? The first byte(s) … Read more
You should read time(7). Be aware that even written in assembler, your program will be rescheduled at arbitrary moments (perhaps a context switch every millisecond; look also into /proc/interrupts and see proc(5)). Then any hardware timer is meaningless. Even using the RDTSC x86-64 machine instruction to read the hardware timestamp counter is useless (since after … Read more
In general, if you use st r1, M1 then value of r1 will be stored into the memory location M1. And in case of st r1, (r3) value of r1 stored into memory of r3 here you have st r1, r2(r3) so value of r1 will be stored into memory of (r2+r3) Correct me if … Read more
I would perhaps start reading about each of those lines one at a time and just see what they do. For example, as someone in the comments said, Read about what int 21 does, it does many things, depending on what is in the AH register. http://www.ctyme.com/intr/int-21.htm e.g. Reading a line from STDIN is specified … Read more
I don’t know if such a book exists (if it does, it’ll likely be a book about compilers). However, there’s an easier solution: try it. Write some C code, then compile it with debug symbols (these instructions assume linux): gcc foo.c -o foo Then, use a debugger: gdb ./foo break MyFunction run disass This will … Read more
This an ancient DOS Protected Mode Interface system call to set a protected mode interrupt vector. See eg http://www.delorie.com/djgpp/doc/dpmi/api/310205.html. The compiler was probably DJGPP. Porting this to a different OS and/or runtime system will require a redesign from scratch to reimplement whatever functionality the interrupt handlers provided under DPMI. Good luck to you with that. … Read more
Jester’s suggestion to write it in C is probably a good one, esp. if it can be inlined into calls where some of the args are compile-time constants. Your example use-case passes mostly compile-time-constant args, including the function pointer. Any decent optimizing compiler will inline this and optimize away the indirection into just a normal … Read more
1.) First of all there is no return statements at the end of the procedure(s). 2.) Since I am using floating point AND STACK I put finit before each procedure call. 3.) I needed to establish a stack frame in the beginning of each procedure push ebp ; save base pointer mov ebp, esp ; … Read more
A loop is implemented like this: loop down: mov ecx, 10 ; count from 10 to 0 label: dec ecx loop label loop up: xor ecx,ecx ; count from 0 to 10 label: inc ecx cmp ecx, 10 jne label Now think how your algorithm works and how to apply this information. 4 solved Division … Read more
(My assembly is a bit rusty, anyone please correct me if I’m wrong): esp is the current stack-pointer, which is where locals and parameters typically live. esp+8 would access an item that is 8 bytes offset from the current stack-frame address. The [x] denotes a dereferencing, so the local is a pointer type. This value … Read more
Your code is being assembled as 16-bit code and you are disassembling it as 32-bit code. solved how to remove null bytes generated at 3rd line in terminal (add [eax],al)?
Sounds like Intel assembly. mov ax, sp; moves the address of the top of the stack into ax. mov ax, [sp]; moves the contents of the top of the stack into ax. 3 solved write the following code (Assembly) [closed]