x86-64

问题 Story Case 1 I accidentally wrote my Assembly code in the .data section. I compiled it and executed it. The program ran normally under Linux 5.4.0-53-generic even though I didn't specify a flag like execstack . Case 2: After that, I executed the program under Linux 5.9.0-050900rc5-generic . The program got SIGSEGV . I inspected the virtual memory permission by reading /proc/$pid/maps . It turned out that the section is not executable. I think there is a configuration on Linux that manages

问题 This question already has answers here : What does the bracket in `movl (%eax), %eax` mean? (3 answers) What does a hexadecimal number, with a register in parenthesis mean in Assembly? (1 answer) Closed 22 days ago . int x=1; int y=2; int z=3; turns into movl $1, -4(%rbp) movl $2, -8(%rbp) movl $3, -12(%rbp) What is the -4,-8,-12 for ? Why is it going by 4's? 4 bytes = 32 bits? 回答1: -4 / -8 / -12 bytes relative to the address held in rbp , which is the pointer to the top of the stack (which

问题 I'm trying to write a procedure in x64 assembly. I'm calling it in a main program that is written in C++. I'm passing several parameters. I know that first 4 will be in specific registers and the rest of them (should be) on stack. What's more, I read that before taking 5th argument from the stack, I should substract 40 from RSP. And at the begining it worked. Later I needed to check the address of sth so I did it by: cout and &. But then, taking 5th argument from stack didn't work and I have

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问题 I followed up this page and compiled the following code ; assembly program that calls a C function on 64-bit Linux ; ; int main(void) { ; printf(fmt, 1, msg1); ; printf(fmt, 2, msg2); ; return 0; ; ; Assemble in 64-bit: nasm -f elf64 -o hp64.o -l hp64.lst hello-printf-64.asm ; ; Link: ld hp64.o -o hp64 -lc --dynamic-linker /lib/ld-2.7.so ; or maybe ld hp64.o -o hp64 -lc --dynamic-linker /lib/ld-linux-x86-64.so.2 ; (the "-lc" option is needed to resolve "printf") ;-----------------------------

问题 I followed up this page and compiled the following code ; assembly program that calls a C function on 64-bit Linux ; ; int main(void) { ; printf(fmt, 1, msg1); ; printf(fmt, 2, msg2); ; return 0; ; ; Assemble in 64-bit: nasm -f elf64 -o hp64.o -l hp64.lst hello-printf-64.asm ; ; Link: ld hp64.o -o hp64 -lc --dynamic-linker /lib/ld-2.7.so ; or maybe ld hp64.o -o hp64 -lc --dynamic-linker /lib/ld-linux-x86-64.so.2 ; (the "-lc" option is needed to resolve "printf") ;-----------------------------

问题 I have the following assembly code .global _start .section .text _start: movq a, %rax movq b, %rbx imul %rbx, %rax cmp %rbx, %rax je gcd_calculated ja l1 sub %rax, %rbx jmp _start l1: sub %rbx, %rax jmp _start gcd_calculated: div %rax movq %rax, (c) a,b are quads that I need to calculate their lcm and I need to assign the result to c I get wrong results with the above code and I can't spot why. generally, i'm relaying on the the lcm = (a*b)/gcd so I store a*b in %rax and then calculate the

问题 I found this page which has a Hello World example for x86-64 on Linux: http://blog.markloiseau.com/2012/05/64-bit-hello-world-in-linux-assembly-nasm/ ; 64-bit "Hello World!" in Linux NASM global _start ; global entry point export for ld section .text _start: ; sys_write(stdout, message, length) mov rax, 1 ; sys_write mov rdi, 1 ; stdout mov rsi, message ; message address mov rdx, length ; message string length syscall ; sys_exit(return_code) mov rax, 60 ; sys_exit mov rdi, 0 ; return 0

问题 I found this page which has a Hello World example for x86-64 on Linux: http://blog.markloiseau.com/2012/05/64-bit-hello-world-in-linux-assembly-nasm/ ; 64-bit "Hello World!" in Linux NASM global _start ; global entry point export for ld section .text _start: ; sys_write(stdout, message, length) mov rax, 1 ; sys_write mov rdi, 1 ; stdout mov rsi, message ; message address mov rdx, length ; message string length syscall ; sys_exit(return_code) mov rax, 60 ; sys_exit mov rdi, 0 ; return 0

问题 Looking at some assembly code for x86_64 on my Mac, I see the following instruction: 48 c7 c0 01 00 00 00 movq $0x1,%rax But nowhere can I find a reference that breaks down the opcode. It seems like 48c7 is a move instruction, c0 defines the %rax register, etc. So, where can I find a reference that tells me all that? I am aware of http://ref.x86asm.net/, but looking at 48 opcodes, I don't see anything that resembles a move. 回答1: Actually, mov is 0xc7 there; 0x48 is, in this case, a long mode