从 C 访问暂存器存储器

发布于 2024-09-10 11:40:26 字数 801 浏览 2 评论 0原文

这可能是一个有点奇怪的问题，但我希望有人可以仍然帮助我一点；）。我想执行一个标准的 C 程序，但是，在程序执行期间的某个时刻，我希望某个特定的存储在本地暂存器 RAM 中的指令数量为被执行。所有进程都可以访问暂存器存储器。让我们假设这个本地内存从地址 0x80000000 开始，我将集成它在下面的 C 代码片段中

int main {
int a=1;
int b=2;
int c=3;    

c = a + b;

%goto address 0x80000000 and execute three instructions before continuing
%program execution here 

return(0);

}

程序计数器将经历以下阶段，假设 main 加载于 0x40000000

0x40000000    a=5; 
0x40000004    b=2; 
0x40000008    c=1;  
0x4000000C    c=a+b;
0x80000000    first instruction in the local scratch pad
0x80000004    second instruction in the local scratch pad
0x80000008    third instruction in the local scratch pad  
0x40000010    return(0);

有人知道如何做到这一点吗？我需要使用汇编跳转吗说明或者有更优雅的东西。

非常感谢，安迪

原文

That might be a little bit an exotic problem, but I hope somebody can
still help me out a bit ;). I would like to execute a standard C program,
however, at some point during program execution I would like that a certain
number of instructions, which are stored in a local scratch pad RAM, are
executed. The scratchpad memory is accessable for all processes. Lets assume
this local memory starts at address 0x80000000 and I would integrate this
in the following C code fragement

int main {
int a=1;
int b=2;
int c=3;    

c = a + b;

%goto address 0x80000000 and execute three instructions before continuing
%program execution here 

return(0);

}

The program counter would go through the following stages, assuming
main is loaded at 0x40000000

0x40000000    a=5; 
0x40000004    b=2; 
0x40000008    c=1;  
0x4000000C    c=a+b;
0x80000000    first instruction in the local scratch pad
0x80000004    second instruction in the local scratch pad
0x80000008    third instruction in the local scratch pad  
0x40000010    return(0);

Anyone an idea how to do this? Do I need to use assembler jump
instructions or is there something more elegant.

Many thanks,
Andi

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

樱桃奶球 2024-09-17 11:40:27

假设这些指令的行为类似于普通函数，您可以这样做：

#include <stdio.h>

void (*scratchpad_func)(void) = (void(*)(void))0x80000000;

int main()
{
    printf("before\n");
    scratchpad_func();
    printf("after\n");
    return 0;
}

显然，您必须使用实模式操作系统，或者跳过操作系统/处理器组合所需的任何障碍才能直接访问该地址空间。

（在某些架构上，如果您不触摸被调用者保存的寄存器，“表现得像一个普通函数”就像最后的“jump $ra”一样简单。例如MIPS。）

Assuming that the instructions behave like a normal function, you can do:

#include <stdio.h>

void (*scratchpad_func)(void) = (void(*)(void))0x80000000;

int main()
{
    printf("before\n");
    scratchpad_func();
    printf("after\n");
    return 0;
}

Obviously, you'll have to be using a real mode operating system, or jump through whatever hoops your OS/processor combination requires to have direct access to that address space.

(On some architectures "behave like a normal function" is as simple as a "jump $ra" at the end if you don't touch callee-saved registers. E.g. MIPS.)

回复收藏 0 原文