在x86中恢复堆栈帧

Question

我正在尝试编写一个将信号实现到xv6中的程序

我已经想出了如何操作堆栈(我认为)，我只是在恢复它时遇到了困难。这是我的信号传送代码：

此功能在处理堆栈中添加信号帧，并保存易失性寄存器。

void signal_deliver(int signum)
{
*((uint*) (proc->tf->esp-4)) = proc->tf->eip;
*((uint*) (proc->tf->esp-8)) = proc->tf->eax;
*((uint*) (proc->tf->esp-12)) = proc->tf->ecx;
*((uint*) (proc->tf->esp-16)) = proc->tf->edx;
*((uint*) (proc->tf->esp-20)) = signum;
*((uint*) (proc->tf->esp-24)) = *(uint*) proc -> signal_trampoline;
proc->tf->esp = proc->tf->esp-24;
proc->tf->eip = (uint) (proc->signal_handlers[signum]);
}

在我的void signal_return(void)中，我在恢复陷阱帧过程时遇到了问题。

我恢复框架的尝试是：

    proc->tf->esp = proc->tf->esp + 24;
    *((uint*)(proc->tf->esp - 16)) = proc->tf->esp;
    *((uint*)(proc->tf->esp - 12)) = proc->tf->esp;
    *((uint*)(proc->tf->esp - 8)) = proc->tf->esp;
    proc->tf->eip = *((uint*)(proc->tf->esp - 4));

谁能给我指明正确的方向？

Answer 1

void signal_return(void) {
    proc->tf->esp = proc->tf->esp + 24;
    proc->tf->edx = *((uint*)(proc->tf->esp - 16));
    proc->tf->ecx = *((uint*)(proc->tf->esp - 12));
    proc->tf->eax = *((uint*)(proc->tf->esp - 8));
    proc->tf->eip = *((uint*)(proc->tf->esp - 4)); 
}