x86的hlt指令不可用

garyv

我在虚拟机里面编译一个c文件(基于x86 gcc)，里面有一条据说是x86的停机指令hlt，如下:

1. int main(...)
   
2. 
   
3. {
   
4. 
   
5.     ......
   
6. 
   
7. 
   
8. 
   
9.     __asm volatile
   
10. 
    
11.     (
    
12. 
    
13.         "hlt"
    
14. 
    
15.     );
    
16. 
    
17. 
    
18. 
    
19.     return 0;
    
20. 
    
21. }
    

复制代码

代码编译正常，但运行时提示"Segmentation fault"，莫非这是一条特权指令？

liuiang

Since issuing the HLT instruction requires ring 0 access, it can only be run by privileged system software, such as the kernel.

cjaizss

这些涉及到硬件特权的指令,当然不可能如此随便

heixia108

搜了一下为什么编译出来的可执行文件有hlt，却要在_exit里退出了

http://stackoverflow.com/questions/5213466/why-does-gcc-place-a-halt-instruction-in-programs-after-the-call-to-main

After main returns, exit will be called. The hlt is there in case the system's version of exit doesn't stop execution of the process immediately. In user mode, it will cause a protection fault, which will kill the process. If the process is for some reason running in ring 0, it will just stop the processor until the next interrupt, which will hopefully trigger the OS to remove the process. In processes designed to run in ring 0, there is often a jmp instruction after the hlt which will cause the hlt to be performed over and over until the process is terminated.

smalloc

查下手册不就知道了
The HLT instruction is a privileged instruction. When the processor is running in
protected or virtual-8086 mode, the privilege level of a program or procedure must
be 0 to execute the HLT instruction.

smalloc

其实HLT的行为就代表不能在用户态执行
An enabled interrupt (including NMI and SMI), a debug exception, the BINIT# signal, the INIT#
signal, or the RESET# signal will resume execution. If an interrupt (including NMI) is
used to resume execution after a HLT instruction, the saved instruction pointer
(CS:EIP) points to the instruction following the HLT instruction.