最短的 GAS ARM (linux) 程序? [英] Shortest possible GAS ARM (linux) program?

问题描述

我萌生了学习汇编语言的念头，并决定尝试一下 ARM.我决定使用 GNU 汇编器，主要是因为它可以在我手机的存储库中找到，这样我就可以在任何地方玩汇编，如果我感到无聊.

I've toyed with the idea of learning an assembly language, and have decided to give ARM a try. I've decided to go with the GNU assembler, mostly because it's available in my cellphone's repository, so that I could play around with assembly anywhere, if I'm bored.

无论如何，我已经在网上搜索过，但找不到任何有关如何正确退出 ARM Linux 二进制文件的参考资料.我了解到 x86 等效项基本上将 eax 寄存器设置为指定系统调用的数字，然后调用系统中断 0x80 以实际执行系统调用，正确退出程序；现在我想为 ARM 做一些类似的事情(显然相同的代码不起作用，因为它使用 x86 特定寄存器等等).

Anyway, I've searched the web, but I can't find any kind of reference for how to properly exit an ARM Linux binary. I've understood that the x86 equivalent basically sets the eax register to a number specifying the system call, and then calls system interrupt 0x80 to actually perform the system call, properly exiting the program; now I want to do something similar for ARM (and obviously the same code doesn't work, since it uses x86 specific registers and whatnot).

所以，是的，基本上，我将如何编写最小的 GAS ARM 可执行文件，只需以退出值 0 正常退出?

So yeah, basically, how would I write a minimal GAS ARM executable, simply exiting normally with exit value 0?

推荐答案

http://www.arm.linux.org.uk/developer/patches/viewpatch.php?id=3105/4.您还可以查看 http://isec.pl/papers/into_my_arms_dsls.pdf 的第 5 页.

There is some information on system calls at http://www.arm.linux.org.uk/developer/patches/viewpatch.php?id=3105/4. You can also look at page 5 of http://isec.pl/papers/into_my_arms_dsls.pdf.

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