Recent program error: Failed to connect to Redis:connection timed outFind information about THP, decide to test and modify1.page and Huge PagesPageIn general, memory management of the minimum block-level units called page, a page is 4096bytes,1m memory will have 256 PAGE,1GB words will have 256,000 page.The CPU maintains the page table record through the built-in memory management unit.Huge Pages:Huge pages is a 2M to 1GB memory page, mainly used to manage thousands of trillion of memory, such a
configuration:Fastcgi_buffers # control Nginx create up to 8 buffers with a size of 4K fastcgi_buffer_size # processing The size of the first buffer when Response is not included in the formerThe maximum memory buffer size that can be created by total is 8*4k+4k = 36kWhen Response is less than or equal to 36k, all of the data is, of course, processed in me
Chengdu 2. What level does 5k PHP programmers need
Like the title--I learned PHP for almost 2 months, has not written a lot of code
Recently learned Html,css, PHP JavaScript then level anyway is not high. And didn't want to get a high salary.
I asked you last time. I am now a 4k C + + QT Programmer
I want to find a PHP job after leaving, but know that their level is not enough to break the study
On the weak weak ask PHP salary structure is what level
also read the original analysis to understand other conclusions that can be obtained.It is not difficult to see that every 1024 steps in the figure, that is 1024*4 , 4096Bytes each, has a particularly obvious blue vertical line. That is, as long as we follow the 4K step into the memory (memory according to 4K alignment), no matter how big the hot data it's actual efficiency is very low! According to our an
~ 1 MB (2^20 * 1 B)
G = 0:4k ~ 4GB (2^20 * 4096 B)
SSytem flag, if the flag is set, indicates that the key data structure, such as the LDT, is stored in the paragraph, otherwise the segment is a normal data segment or code snippet.
TYPEIndicates the type of the segment, as well as access rights.
DplDescriptor Privilege Level: Used to restrict access to this segment, indicating the minimum permissions required to access the seg
[i]=1, which is marked as occupied by the page, not the idle bit(5) "Sall $12,%%ecx\n\t"At this time ecx save is mem_map[i] subscript I, that is, the relative number of pages,Example:Suppose Mem_map[0: ( PAGING_PAGES-1)] [last parameter]MEM_MAP[PAGING_PAGES-1] = = 0, i.e. i = = (paging_pages-1),So at this time *ecx = = paging_pages-1;At this point the relative page address is 4k* (paging_pages-1),1024 4-byte physical pages per page, 12-bit left shift
is mem_map[i] subscript I, that is, the relative number of pages,Example:Suppose Mem_map[0: ( PAGING_PAGES-1)] [last parameter]MEM_MAP[PAGING_PAGES-1] = = 0, i.e. i = = (paging_pages-1),So at this time *ecx = = paging_pages-1;At this point the relative page address is 4k* (paging_pages-1),1024 4-byte physical pages per page, 12-bit left shift equals 4096 (2 12-square),(6) "Addl%2,%%ecx\n\t"Add a low-end memory address to get the actual physical addre
is likely due to high CPU utilization, rather than high CPU utilization due to high running queues. The number of active sessions is not particularly high from the database. So next, do you need to focus on whether the system memory management issues are causing high CPU utilization?Reviewing the system memory aspects of/proc/meminfo collected at the beginning of this article, you can find an important piece of data:
pagetables:4749076 KB
As you can see from the data, the Pagetabl
size can be specified before formatting. But generally not to change. To view the inode size command:650) this.width=650; "src=" Http://s5.51cto.com/wyfs02/M02/86/36/wKiom1e4MwuQFkyTAABDr99Vln4130.png "title=" Snap28.png "alt=" Wkiom1e4mwuqfkytaabdr99vln4130.png "/>A file occupies at least one inode and one block. Block is where the internship holds the content. Block large, will waste disk space, small, affecting disk read performance, general production environment set block size of 4KAsk 100
Each process has its own page table, in Linux, there is a page directory array, which is the top level of the paging mechanism, and each process's page table corresponds to one of the page catalog entries, which can be accessed through the CR3 register.A Process page table that corresponds to the physical address of the page in the page table entry.2. Paging mechanismThe role of paging mechanismThe paging mechanism is performed after the segment mechanism, which further translates the linear add
;
# Disables the gzip feature when the client is IE6.
Gzip_disable "Msie6"; #Nginx做为反向代理的时候启用.
Optional value: Off|expired|no-cache|no-sotre|private|no_last_modified|no_etag|auth|any gzip_proxied any; #设置允许压缩的页面最小字节数, the number of page bytes is fetched from the Content-length in the header header.
It is recommended that you set the number of bytes larger than 1k, less than 1k may be more pressing.
Gzip_min_length 1024; #设置数据的压缩等级.
This level can be any number between 1-9, and
1, the U start installed version of the U disk plug in the computer, and then restart the computer, in the boot screen with a key U-disk to start shortcut keys to start the main menu interface, select "03" Run U start win03pe enhanced version, press ENTER to perform
2, double hit open Diskgenius partition tool, in the Pop-up tool window click to select the partition, and then click the "Delete Partition" button
3, at this point will pop up a prompt to do this will delet
Let's take a look at the details of the U disk:
The above is the chip worry-free detection of my U disk information, my U disk capacity of 16GB, total flash memory is also 16GB, the description has not been expanded, is the real 16GB. This is a big thing to worry about. So what's the reason? Let's take a look at it again:
Notice that the partition system default is generally Fat32,fat32 is Microsoft's very early file system, but also commonly used in mobile
How to install Win7 on solid State drive
I. Installation and preparation
1, solid State hard drive installed on the computer
2, 4G u disk: Making u disk PE boot disk
3, operating system: Win7 64-bit flagship edition
4, startup settings: Set boot from U disk Boot method
5, 4K Alignment Tutorial: SSD Solid State hard disk 4k alignment tutorial
Second, solid state hard drive mounted Win7 steps as follo
. The directio_alignment size depends on the file system you use, the default is 512, and for XFS, note that if you do not modify the XFS bsize, you need to adjust to the XFS default 4k.
The configuration I used is as follows:
Copy Code code as follows:
location/archive {
Internal
AIO on;
Directio 4k;
Directio_alignment 4k;
Output_buffer
is based on the page concept, for example, OS VM1 page is 4K, 4K as long as there is only one element even if only 1 bytes are accessed, the page will not be swap, the same reason, read a byte may be swapped into 4K useless memory. and Redis own implementation can achieve control of the size of the swap. The block process in addition to accessing the operating s
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