问题描述
尝试运行以下汇编程序时：
When attempting to run the following assembly program:
.globl start
start:
pushq $0x0
movq $0x1, %rax
subq $0x8, %rsp
int $0x80
我收到以下错误：
dyld: no writable segment
Trace/BPT trap
任何人都知道是什么原因引起的吗？
Any idea what could be causing this? The analogous program in 32 bit assembly runs fine.
推荐答案
OSX现在要求您的可执行文件具有包含内容的可写数据段，因此，该程序运行良好。它可以动态地重新定位和链接您的代码。 Dunno为什么，也许是安全原因，也许是由于新的RIP寄存器。如果在其中放置.data段（内容有些虚假），则可以避免出现 no writable segment错误。 IMO，这是一个ld bug。
OSX now requires your executable to have a writable data segment with content, so it can relocate and link your code dynamically. Dunno why, maybe security reasons, maybe due to the new RIP register. If you put a .data segment in there (with some bogus content), you'll avoid the "no writable segment" error. IMO this is an ld bug.
对于64位系统调用，您可以使用两种方法来实现。 GCC样式，它使用来自libSystem.dylib的_syscall PROCEDURE或raw。 Raw使用syscall指令，而不是int 0x80陷阱。 int 0x80是64位上的非法指令。
Regarding the 64-bit syscall, you can do it 2 ways. GCC-style, which uses the _syscall PROCEDURE from libSystem.dylib, or raw. Raw uses the syscall instruction, not the int 0x80 trap. "int 0x80" is an illegal instruction in 64-bit.
GCC方法将为您对syscall进行分类，因此您可以使用相同的32位sys / syscall.h中找到的位数。但是，如果您一无所获，则必须通过将其与类型ID进行或运算来对它进行分类。这是两个例子。请注意，调用约定是不同的！ （对NASM语法表示歉意；气体使我烦恼）
The "GCC method" will take care of categorizing the syscall for you, so you can use the same 32-bit numbers found in sys/syscall.h. But if you go raw, you'll have to classify what kind of syscall it is by ORing it with a type id. Here is an example of both. Note that the calling convention is different! (apologies for NASM syntax; gas annoys me)
; assemble with
; nasm -f macho64 -o syscall64.o syscall64.asm && ld -lc -ldylib1.o -e start -o syscall64 syscall64.o
extern _syscall
global start
[section .text align=16]
start:
; do it gcc-style
mov rdi, 0x4 ; sys_write
mov rsi, 1 ; file descriptor
mov rdx, hello
mov rcx, size
call _syscall ; we're calling a procedure, not trapping.
;now let's do it raw
mov rax, 0x2000001 ; SYS_exit = 1 and is type 2 (bsd call)
mov rdi, 0 ; Exit success = 0
syscall ; faster than int 0x80, and legal!
[section .data align=16]
hello: db "hello 64-bit syscall!", 0x0a
size: equ $-hello
签出有关如何键入系统调用的更多信息。
check out http://www.opensource.apple.com/source/xnu/xnu-792.13.8/osfmk/mach/i386/syscall_sw.h for more info on how a syscall is typed.
这篇关于Mac OS X运行时错误的64位汇编：“ dyld：no writable segment”和“跟踪/ BPT陷阱”；的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！