分配最大的缓冲而不使用交换 [英] Allocating largest buffer without using swap

问题描述

在Linux下的C / C ++，我需要分配一个大的内存（数千兆字节）块，以从连接到以太网端口和流数据大约110MB / s的传感器，存储实时数据。我想分配的内存可能量最大，以最大限度地提高，我可以存储数据序列的长度。然而，我还需要确保不会有圆盘交换，因为所得延迟和磁盘访问有限带宽引起传感器的（非常有限的）缓冲器溢出。

In C/C++ under Linux, I need to allocate a large (several gigabyte) block of memory, in order to store real-time data from a sensor connected to the ethernet port and streaming data at about 110MB/s. I'd like to allocate the largest amount of memory possible, to maximise the length of data sequence that I can store. However, I also need to make sure that there will be no disk-swapping, since the resulting delay and limited bandwidth of disk access causes the sensor's (very limited) buffer to overflow.

什么是确定有多少内存分配的最佳方式？难道我只限于分配大于报告的可用内存稍小块，或者我可以更直接地与Linux虚拟内存管理器界面？

What is the best way to determine how much memory to allocate? Am I limited to just allocating a slightly smaller block than the reported free memory, or can I interface more directly with the linux virtual memory manager?

推荐答案

那么，在Linux下你可以使用MLOCK（）/ mlockall用于（），以保持在物理内存和prevent它的ADRESS范围内被交换出去。使用MLOCK这个过程需要一对夫妇的权限这样做，人MLOCK有详细信息。我不知道有关最高mlock'able块（它可能会从什么似乎是免费的不同），所以可能是一个二进制搜索可以帮助（锁定一个范围，如果失败减少面积等尺寸。）

Well, under linux you can use mlock()/mlockall() to keep an adress range in physical memory and prevent it from being swapped out. The process using mlock needs a couple of privileges to do so, "man mlock" has the details. I am not sure about the maximum mlock'able block (it might differ from what seems to be "free"), so probably a binary search could help (lock a range, if that fails reduce the size of the area etc..)

在另一方面，110MB / s是不是真的固态驱动器有问题。 60GB的SSD具有280MB / s的写入速度在角落里的成本约为$ 200。只是将传感器数据复制到一个小的写缓冲器和流是到SSD。

On the other hand, 110MB/s is not really a problem for a Solid-State-Drive. A 60GB SSD with 280MB/s write speed costs about $200 on the corner. Just copy the sensor data into a small write buffer and stream that to the SSD.

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