无法使用ld链接目标文件 - Mac OS X.

/*********
exit.asm
*/

[SECTION .text]

global _start


_start:
xor eax, eax
xor ebx, ebx
mov al, 1
int 0x80

//****************************

First I used nasm -f elf exit.asm to generate the object file.

首先，我使用nasm -f elf exit.asm生成目标文件。

then I ran the following "ld" command on my Mac OS X 10.7, it has the these outputs and warning, I tried to run it on my 32 bit linux machine, everything went through just fine, Could you please explain why would not the linker work on my Mac?

然后我在我的Mac OS X 10.7上运行了以下“ld”命令，它有这些输出和警告，我试图在我的32位linux机器上运行它，一切都很顺利，请你解释为什么不会链接器在我的Mac上运行？

Thank you!

谢谢！

Alfred says: ld -o exiter exit.o
ld: warning: -arch not specified
ld: warning: -macosx_version_min not specified, assuming 10.7
ld: warning: ignoring file exit.o, file was built for unsupported file format ( 0x7f 0x45      0x4c 0x46 0x 1 0x 1 0x 1 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 ) which is not the   architecture being linked (x86_64): exit.o
Undefined symbols for architecture x86_64:
  "start", referenced from:
    implicit entry/start for main executable
ld: symbol(s) not found for inferred architecture x86_64

after I specify my arch and version, I got:

在我指定我的拱门和版本之后，我得到了：

Alfred says: ld -arch x86_64 -macosx_version_min 10.7 -o exiter exit.o
ld: warning: ignoring file exit.o, file was built for unsupported file format ( 0x7f 0x45     0x4c 0x46 0x 1 0x 1 0x 1 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 0x 0 ) which is not the   architecture being linked (x86_64): exit.o
 Undefined symbols for architecture x86_64:
 "start", referenced from:
    implicit entry/start for main executable
 ld: symbol(s) not found for architecture x86_64

4 个解决方案

#1

Getting the program to link is the easy part:

获取程序链接很容易：

• Change _start to start
• 改变_start开始
• $ nasm -f macho exit.asm
• $ nasm -f macho exit.asm
• $ ld -arch i386 -o exiter exit.o
• $ ld -arch i386 -o exiter exit.o

The problem is that exit.asm is calling the i386 Linux exit() system call (EAX = 1) and the program would NOT exit with a zero status as intended on OS X.

问题是exit.asm正在调用i386 Linux exit（）系统调用（EAX = 1），并且程序不会以OS X上的预期状态退出。

System Calls

A system call is a request to the kernel. exit(), unlike sqrt(), must make a request to a software component with higher privileges in its implementation since it terminates a running program. Apps can't create or terminate processes by themselves. System calls provide a way for apps to ask the kernel to perform actions on their behalf.

系统调用是对内核的请求。与sqrt（）不同，exit（）必须向其实现中具有更高权限的软件组件发出请求，因为它终止正在运行的程序。应用无法自行创建或终止进程。系统调用为应用程序提供了一种让内核代表他们执行操作的方法。

Making a syscall goes something like this:

制作系统调用是这样的：

• Apps describe the operation they want to perform by placing data in CPU registers (or memory pointed to by registers), e.g.
  • The value 1 in EAX is the system call number of exit.
  • EAX中的值1是退出的系统调用号。
  • The value 0 in EBX (EBX was cleared by xor) is the first argument to the syscall, the exit status.
  • EBX中的值0（EBX由xor清除）是系统调用的第一个参数，即退出状态。
• 应用程序通过将数据放入CPU寄存器（或寄存器指向的存储器）来描述他们想要执行的操作，例如， EAX中的值1是退出的系统调用号。 EBX中的值0（EBX由xor清除）是系统调用的第一个参数，即退出状态。
• Apps issue an instruction that causes control to transfer to the kernel, e.g.
  • int 80 on i386
  • i386上的int 80
  • sycall on x86-64
  • x86-64上的sycall
  • svc in Thumb mode on ARMv7
  • ARMv7上的Thumb模式下的svc
• 应用程序发出一条指令，使控件转移到内核，例如在ARMv7上的Thumb模式下，x86-64 svc上的i386 sycall上的int 80
• The kernel inspects the request and decides to perform or deny it.
• 内核检查请求并决定执行或拒绝它。
• The kernel transfers control back to the app with the return value in an agreed upon location, e.g. EAX on i386.
• 内核将控制权转移回应用程序，并在约定的位置返回值，例如： i386上的EAX。

Linux and OS X both provide a void exit(int) function for C programs but don't agree on the details on how to describe this request to the kernel. The code in exit.asm is at the same level as the implementation of the _exit() function in libc.

Linux和OS X都为C程序提供了一个void exit（int）函数，但是对于如何向内核描述这个请求的细节并不一致。 exit.asm中的代码与libc中_exit（）函数的实现处于同一级别。

Even between different architectures running Linux the syscall numbers and calling convention differ. e.g. On x86-64 Linux, exit(0) is more commonly issued like this:

即使在运行Linux的不同架构之间，系统调用号和调用约定也不同。例如在x86-64 Linux上，退出（0）更常发布如下：

xor rdi, rdi
mov al, 60
syscall

You can see this by disassembling _exit in /lib64/libc.so.6.

你可以通过在/lib64/libc.so.6中反汇编_exit来看到这一点。

Can't We Just Call exit() from libc Instead?

You can. But you'd have to link the program with libc. It's the difference between linking exit.asm above with:

您可以。但是你必须将程序与libc链接起来。链接exit.asm与上面的区别是：

$ cc -m32 -nostdlib exit.o -o exiter

and

和

exit-libc.asm

出口libc.asm

extern exit
global main
main:
push 0
call exit

which has to be linked with:

必须与以下内容联系：

$ cc -m32 exit-libc.o -o exit-libc

Try this and take a look at the file size.

试试这个，看看文件大小。

#2

Mac OS X doesn't use ELF, so you'll want to generate a Mach-O object to link with on that system. On my machine nasm appears to only support 32-bit output, so you'll need to match that architecture when linking, too.

Mac OS X不使用ELF，因此您需要生成一个Mach-O对象以在该系统上进行链接。在我的机器上，nasm似乎只支持32位输出，因此在链接时你也需要匹配该架构。

I also had to change _start to start to get it to link.

我还必须更改_start以开始链接。

Here's a working example with your code:

以下是您的代码的工作示例：

$ cat exit.asm 
[SECTION .text]

global start

start:
xor eax, eax
xor ebx, ebx
mov al, 1
int 0x80
$ nasm -f macho exit.asm 
$ ld -arch i386 -macosx_version_min 10.7  -o exiter exit.o 
$ ./exiter 
$ echo $?
236

Note that the program probably doesn't do what you want on Mac OS X, since it doesn't do system calls the same way as Linux.

请注意，该程序可能无法在Mac OS X上执行您想要的操作，因为它不像Linux那样执行系统调用。

#3

Most of the time when you receive this error:

大多数情况下，当您收到此错误时：

ld: warning: PIE disabled. Absolute addressing (perhaps -mdynamic-no-pic) not 
allowed in code signed PIE, but used in _start from hello.o. To fix this 
warning, don't compile with -mdynamic-no-pic or link with -Wl,-no_pie

It is because it is looking for your "main()" function(label) to me the label "start:". It is always best to specify the your main label with "ld -e".

这是因为它正在寻找你的“main（）”函数（标签）给我标签“start：”。最好用“ld -e”指定主标签。

For nasm:

-o hello.tmp - outfile
-f macho - specify format
Linux - elf or elf64
Mac OSX - macho

For ld:

-arch i386 - specify architecture (32 bit assembly)
-macosx_version_min 10.6 (Mac OSX - complains about default specification)
-no_pie (Mac OSX - removes ld warning)
-e main - specify main symbol name (Mac OSX - default is start)
-o hello.o - outfile

For Shell:

./hello.o - execution

One-liner:

nasm -o hello.tmp -f macho hello.s && ld -arch i386 -macosx_version_min 10.6 -no_pie -e _main -o hello.o hello.tmp && ./hello.o

Let me know if this helps!

如果这有帮助，请告诉我！

I wrote how to do it on my blog here:

我在博客上写了如何做到这一点：

http://blog.burrowsapps.com/2013/07/how-to-compile-helloworld-in-intel-x86.html

http://blog.burrowsapps.com/2013/07/how-to-compile-helloworld-in-intel-x86.html

For a more verbose explanation, I explained on my Github here:

有关更详细的解释，我在这里向我的Github解释：

https://github.com/jaredsburrows/Assembly

https://github.com/jaredsburrows/Assembly

#4

The standard mac gcc won't link elf objects. For people who need to stick with the elf format and develop on a mac, you need a cross compiler...

标准mac gcc不会链接elf对象。对于需要坚持使用elf格式并在mac上开发的人来说，你需要一个交叉编译器......

http://crossgcc.rts-software.org/doku.php?id=compiling_for_linux

http://crossgcc.rts-software.org/doku.php?id=compiling_for_linux

Then you can proceed with something similar to this...

然后你可以继续进行类似的事情......

/usr/local/gcc-4.8.1-for-linux32/bin/i586-pc-linux-ld -m elf_i386 -T link.ld -o kernel kasm.o kc.o