在C或C++不使用某种种类的议会(最好不使用__asm)的情况下,能否创建一个箱子? 我撰写一个操作系统,希望在C和C++中填写。
That s pretty system dependent. In most cases, the answer is going to be no - you will need to write some custom assembly to set up the C runtime before you start running your C code. There are some exceptions, however. The ARM Cortex-M0, for example, can run C code straight out of reset.
简言之,你不再使用M0,因此,你需要写一些集会。 同样,它依赖系统/芯片,但你可能能够放弃简单的东西:
reset_vector:
mov sp, SOME_KNOWN_GOOD_STACK_ADDRESS
call c_entry_point
这只是先入住脚点,把我们的C方案放在切入点。 当然,这种简单的设置取决于您的心灵,即有一个支持它的复发病媒/病媒表,而区域援助团(或像区域援助团)是在重新启动病媒之前启动的,等等。 早期系统启动时往往有许多“障碍”。
Prepare to get pretty friendly with your compiler, assembler, and linker documentation - generating a flat binary that you can flash down as a first stage bootloader is often a big pain in and of itself.
亲爱!
假设是x86,如果你有正确的编造者选择,并设法使boot部门的布局与适当的连接者魔法保持一致,那么你可能获得16个轨道模式的something。
但是,你赢得了远近的光线(C(或C++):你需要掩饰真正的快速中断,因此没有<代码>C功能。 需要议会。
对大多数其他结构来说,这大概相同:C和C++根本不具备这些特征(有些编汇的延伸可能有助于你)。
http://wiki.osdev.org/“rel=“nofollow”>OSDev。
我认为答案是No,因为你需要做这样的事情才能转向受保护的方式:
lgdt[GDTR]
jmp CODESEL:FLUSH
FLUSH:
...
我不认为,在纯粹的C/C++中采用“<>jmp指示的方法,尽管我可能错。 (我绝不是这里的一位专家;我只提一下我以前曾做过过的boot装机。)
否,它不可能有“纯”C,至少是x86......。 除了X86机器将安装在16轨实时(要求汇编者生成16比而不是32比方代码)之外,你还需要能够掩饰打断、编组部分登记册、从硬件装置(即软盘)中装载密码、设置一个排 st、进入I/O港等,所有这些平台都要求在x86和其他平台上查阅CPU的登记册和(或)特定组装指挥部。
第二,对于C++来说,如果你决定确定任何类别,那么你将需要有某种人工配置和操作的“建筑”,以积累记忆,使你最初的班级实际上能够生存......你也获得了任何例外。 基本上,你试图使用的任何C++具体特征都将是无用的,因为这些高层次的数据记录需要本组织日常运作本身的适当支持,而这种支持必须使用组合和C编码加以设置。
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