gcc - 我如何进一步了解 gcc 使用的编译过程？

Question

我试图对一些使用免费 pspsdk 开发的 psp 程序进行逆向工程

https://sourceforge.net/projects/minpspw/

我注意到我创建了一个函数来查看 MIPS 如何处理超过 4 个参数 (a0-a4)。我认识的每个人都告诉我他们被传递到堆栈上。令我惊讶的是，第 5 个参数实际上被传递给寄存器 t0 并且编译器甚至没有使用堆栈！

它还内联了一个函数，甚至没有使用 jal 或跳转到它。（明显的优化）。尽管确实有一个内存空间，您可以通过使用带有函数指针参数的 print 来仔细检查。执行的实际代码是自动内联的，不需要函数调用指令。

^^ 但这对于逆向工程师的尝试并没有真正让我受益......

这个版本的 gcc 有一个手册页。如果有人能够提供编译的人（如果有的话），安装需要几秒钟。太长了，我什至不知道如何可靠地参考信息

标签： gcccompilation

Answer 1

如何传递参数由 ABI（应用程序二进制接口）指定。所以你必须找到相应的文件。

此外，还有不止一种这样的 ABI，即n32和n64。在.mips-gcc _

/* This structure has to cope with two different argument allocation
   schemes.  Most MIPS ABIs view the arguments as a structure, of which
   the first N words go in registers and the rest go on the stack.  If I
   < N, the Ith word might go in Ith integer argument register or in a
   floating-point register.  For these ABIs, we only need to remember
   the offset of the current argument into the structure.

   The EABI instead allocates the integer and floating-point arguments
   separately.  The first N words of FP arguments go in FP registers,
   the rest go on the stack.  Likewise, the first N words of the other
   arguments go in integer registers, and the rest go on the stack.  We
   need to maintain three counts: the number of integer registers used,
   the number of floating-point registers used, and the number of words
   passed on the stack.

   We could keep separate information for the two ABIs (a word count for
   the standard ABIs, and three separate counts for the EABI).  But it
   seems simpler to view the standard ABIs as forms of EABI that do not
   allocate floating-point registers.

   So for the standard ABIs, the first N words are allocated to integer
   registers, and mips_function_arg decides on an argument-by-argument
   basis whether that argument should really go in an integer register,
   or in a floating-point one.  */

mips 后端有更多这样的评论。mips.c在和中搜索“累积”或“累积” mips.h。