谁能从 "stack" 的角度解释这段汇编代码？

Question

任何人都可以从堆栈的角度解释基于 arm 的汇编代码吗？特别是在调用“main”、“save_context”和“resume”部分之前“reset_handler”的堆栈视图？（请注意，我知道代码在做什么，但我无法理解或想象代码在运行时堆栈的外观或行为如何）。

   */ asm.s */

  .global main, process, process_size
  .global reset_handler, context_switch, running

reset_handler:
   ldr r0, =process
   ldr r1, =process_size 
   ldr r2, [r1, #0] 
   add r0, r0, r2 
   mov sp, r0 
   bl main 
   
context_switch:
 save_context:
  stmfd sp!, {r0-r12, lr}
  ldr r0, =running 
  ldr r1, [r0, #0] 
  str sp, [r1, #4] 

 resume:
  ldr r0, =running
  ldr r1, [r0, #0] 
  ldr sp, [r1, #4] 
  ldmfd sp!, {r0-r12, lr} 
  mov pc, lr

*/ cfile.c */

#define SIZE 2048 
typedef struct process
{
   struct process *next; 
   int *saved_stack;
   int running_stack[SIZE]; 
}PROC;

int process_size = sizeof(PROC);

PROC process, *running; 

main() 
{
  running = &process; 
  context_switch();
}

Answer 1

伪代码：

reset_handler:
  stack_pointer = &process+process_size;
  call main

这意味着堆栈指针指向&process.running_stack[SIZE]。当一个栈指针像这样指向栈缓冲区的末尾时，就意味着栈是完全空的（ARM使用递减栈）。

Answer 2

作为背景——处理器的寄存器几乎定义了它在做什么。它们通常被称为 context。最重要的是程序计数器pc，它包含下一条指令的内存地址；但是它们都很重要。那么让我们看看如何保存上下文：

save_context:
  stmfd sp!, {r0-r12, lr}
     -- that instruction saved to processor context to the stack
     -- it could be broken down as follows:
     -- sp = sp - 14*4    4, because each register is 4 bytes, and there are 14 specfied
     -- for (i=0; i < 13; i++)   sp[i] = r(i);
     -- sp[i] = lr      `lr` is special, it holds the return address of the instruction that called us.
  ldr r0, =running 
     -- put the address of the variable `running` into r0
  ldr r1, [r0, #0] 
     -- load r1 with the memory address from r0.  So r1 = running.
  str sp, [r1, #4] 
     -- store the stack pointer (sp) in the `saved_sp` field of running.
     -- so these three instructions perform:  running->saved_stack = sp;
     -- now we "fall through" to load, or `resume` a context.
 resume:
  ldr r0, =running
  ldr r1, [r0, #0] 
  ldr sp, [r1, #4]
      -- the inverse of the above, these three instructions effectively perform:
      -- sp = running->saved_stack 
  ldmfd sp!, {r0-r12, lr} 
      -- this is the complimentary operation to the complicated save one above; but this time it is:
      -- for (i=0; i < 13; i++) r(i) = sp[i];
      -- lr = sp[i];
      -- sp += 14*4;
  mov pc, lr
      -- this is a return instruction, where the program counter is loaded with the contents of the link register `lr`.
      -- so, with this, it will return to main just after the call to context_switch

上面有一些模糊的地方：sp[i] 必须按寄存器的大小缩放 i (4)；但较早的 sp 减少了 14*4。由于伪C不是真的，所以看起来还可以。

谁能从 "stack" 的角度解释这段汇编代码？

Can anyone explain this assembly code for the "stack" point of view?

multithreading

assembly

process

operating-system

context-switching