我正在尝试在OSX上的汇编中编写x86-64 hello世界,但每当我编写一个系统调用时,它都是segfaulting。我已经通过Gnu C内联汇编尝试了相同的系统调用并且它可以工作,所以我很困惑:
.section __TEXT,__text,regular,pure_instructions
.globl _main
.align 4, 0x90
_main:
.cfi_startproc
movq 0x2000004, %rax
movq 1, %rdi
leaq _hi(%rip), %rsi
movq 12, %rdx
syscall
xor %rax, %rax
ret
.cfi_endproc
.section __DATA,__data
.globl _hi
_hi:
.asciz "Hello world\n"
这是基于以下Gnu C,它起作用:
#include <string.h>
int main() {
char *hw = "Hello World\n";
unsigned long long result;
asm volatile ("movq %1, %%rax\n"
"movq %2, %%rdi\n"
"movq %3, %%rsi\n"
"movq %4, %%rdx\n"
"syscall\n"
: "=rax" (result)
: "Z" (0x2000004),
"Z" (1),
"r" (hw),
"Z" (12)
: "rax", "rdi", "rsi", "rdx");
}
编译时的C块生成以下asm:
.section __TEXT,__text,regular,pure_instructions
.globl _main
.align 4, 0x90
_main: ## @main
.cfi_startproc
## BB#0:
pushq %rbp
Ltmp2:
.cfi_def_cfa_offset 16
Ltmp3:
.cfi_offset %rbp, -16
movq %rsp, %rbp
Ltmp4:
.cfi_def_cfa_register %rbp
leaq L_.str(%rip), %rcx
movq %rcx, -8(%rbp)
## InlineAsm Start
movq $33554436, %rax
movq $1, %rdi
movq %rcx, %rsi
movq $12, %rdx
syscall
## InlineAsm End
movq %rcx, -16(%rbp)
xorl %eax, %eax
popq %rbp
ret
.cfi_endproc
.section __TEXT,__cstring,cstring_literals
L_.str: ## @.str
.asciz "Hello World\n"
答案 0 :(得分:0)
在这种情况下,这是一个简单的64位“Hello World”(或Hello StackOverflow)。它应该建立在OSX上。试一试:
section .data
string1 db 0xa, " Hello StackOverflow!!!", 0xa, 0xa, 0
len equ $ - string1
section .text
global _start
_start:
; write string to stdout
mov rax, 1 ; set write to command
mov rsi, string1 ; string1 to source index
mov rdi, rax ; set destination index to 1 (stdout) already in rax
mov rdx, len ; set length in rdx
syscall ; call kernel
; exit
xor rdi,rdi ; zero rdi (rdi hold return value)
mov rax, 0x3c ; set syscall number to 60 (0x3c hex)
syscall ; call kernel
; **Compile/Output**
;
; $ nasm -felf64 -o hello-stack_64.o hello-stack_64.asm
; $ ld -o hello-stack_64 hello-stack_64.o
; $ ./hello-stack_64
;
; Hello StackOverflow!!!