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POHow do you understand mlockall man page?
 

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  1. POHow do you understand mlockall man page?
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    <p>The man page for <code>mlockall</code> on my kernel 3.0 says</p> <blockquote> <p>mlockall() locks all pages mapped into the address space of the calling process. This includes the pages of the code, data and <strong>stack</strong> segment, as well as shared libraries, user space kernel data, shared memory, and memory-mapped files. All mapped pages are guaranteed to be resident in RAM when the call returns successfully; the pages are guaranteed to stay in RAM until later unlocked.</p> </blockquote> <p>and later says</p> <blockquote> <p>Real-time processes that are using mlockall() to prevent delays on page faults should reserve enough locked stack pages before entering the time-critical section, so that no page fault can be caused by function calls. This can be achieved by calling a function that allocates a sufficiently large automatic variable (an array) and writes to the memory occupied by this array in order to touch these stack pages. This way, enough pages will be mapped for the stack and can be locked into RAM. The dummy writes ensure that not even copy-on-write page faults can occur in the critical section.</p> </blockquote> <p>I understand that this system call can't guess the maximum stack size that will be reached and thus is unable to lock pages for the stack. But why the first part of the man displayed above says that it's also done for the stack ? Is there an error in this man page, or does it just mean that the locking is done for the initial stack size ?</p>
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