我可以告诉Linux不要换出特定进程的内存吗? [英] Can I tell Linux not to swap out a particular processes' memory?

问题描述

有没有办法告诉Linux，它不应该将特定进程的内存换出到磁盘上?

Is there a way to tell Linux that it shouldn't swap out a particular processes' memory to disk?

它是一个Java应用程序，因此理想情况下，我希望从命令行中执行此操作.

Its a Java app, so ideally I'm hoping for a way to do this from the command line.

我知道您可以将全局swappiness设置为0，但这是明智的吗?

I'm aware that you can set the global swappiness to 0, but is this wise?

推荐答案

您可以通过 mlockall(2)系统调用；这将适用于整个过程，但请务必阅读您需要传递的参数.

You can do this via the mlockall(2) system call under Linux; this will work for the whole process, but do read about the argument you need to pass.

您真的需要将所有内容都放入核心吗?如果它是一个Java应用程序，则可能会将整个JVM锁定在内核中.我不知道执行此操作的命令行方法，但是您可以编写一个简单的程序来依次调用fork，先调用mlockall然后再调用exec.

Do you really need to pull the whole thing in-core? If it's a java app, you would presumably lock the whole JVM in-core. I don't know of a command-line method for doing this, but you could write a trivial program to call fork, call mlockall, then exec.

您还可能希望查看 madvise(2)满足您的需求.如果VM子系统适合您，则向其建议有关更好的分页策略的建议可能会更好.

You might also look to see if one of the access pattern notifications in madvise(2) meets your needs. Advising the VM subsystem about a better paging strategy might work out better if it's applicable for you.

请注意，很久以前，在SunOS下，存在一种类似于madvise的机制，称为

Note that a long time ago now under SunOS, there was a mechanism similar to madvise called vadvise(2).

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