What exactly happens inside the OS to cause the segmentation fault

折月煮酒 提交于 2019-12-21 06:25:49

问题


I have read a lot about the virtual address and paging. Let me first tell you people what i understood. When a process wants to execute something it tries to load the data from the hard disk to memory. To do this it uses a virtual address. So our MMU validates the virtual address looks into the TLB to find the corresponding physical page, if it doesn't find there it looks into Inverted Page Table and at the end it looks into Page table if it doesn't find an entry over there it generates a page fault and all the swapping of page is done and all the tables will be updated. And as I read all the processes have different page tables and same virtual address. so if I try access an array element a[1000] which was defined as int a[100] I am sure that am gonna get a segmentation fault cause that instruction might be trying to access a memory that doesn't belong to it. but how OS comes to know that a[1000] doesn't belong to the running process by just using the concept of virtual address and physical pages. Am I missing something here or my entire understanding is wrong?

I know we can say a memory access is illegal if a process is trying to access a read only or sup true memory segment.

at the end the boiling question is how OS decides which memory is allocated to which process and how it decides that this access of memory is illegal.

What is a segmentation fault on Linux?

this link didn't help much .

thanks a lot in advance for all you lovely people's inputs :)


回答1:


Actually each and every process in the Linux kernel has some well defined structure associated with it called task_struct which stores all the information about the corresponding process like its parent, its PID, its child, its address space, pending signals, threads associated with it etc. Now the address space entries tell the kernel while executing the process the legal address space for that process.Every process has its own address space allotted to it by the kernel right from the very beginning when it is created. So when a executing process tries to access the space outside its legal memory a fault is generated(called segmentation fault in Unix/Linux) and the process is terminated by giving a signal to it by the kernel. Its important for memory protection to be achieved in OS.




回答2:


On x86, linux uses a combination of segmentation and paging, so the address generated by program first looks up for the corresponding segments base and limit registers values. This gives the virtual address which is then translated using the page table. When you try to access a memory which has not been allocated, the accessed page is beyond what the limit register allows, hence generating a segmentation fault.



来源:https://stackoverflow.com/questions/13225936/what-exactly-happens-inside-the-os-to-cause-the-segmentation-fault

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