linux network utilization

TEST_USLEEP.O6. #define ELAPSEDTIME 10000//10000 μs =10 msCPU usage:0.3%%cpu%MEM time+963403368868 760 0.3 0.0 0:00.047. #define ELAPSEDTIME 100000//100000 μs =100 msCPU usage:0.0%%cpu%MEM time+964403368872 760 0.0 0.0 0:00.00 TEST_USLEEP.OIv. Reference Information1.linux, get CPU informationCat/proc/cpuinfo2.linux, get memory informationCat/proc/meminfo3.linux,

, where the number of users will also appear as the number of terminals; "Load average" is the average of the current system load, the following three values are 1 minutes ago, 5 minutes ago, 15 minutes before the average process, the general can be considered that the number of CPUs, the CPU will be more laborious load the current system contains processes; Second line (Tasks): "Total" is the current number of system processes; "1 Running" is the number of processes currently running; "Sleeping

Proc File System IntroductionThe/proc file system is a pseudo-file system that only exists in memory and does not occupy external memory space. It provides the interface for the kernel to communicate with the process in a file system manner. Users and applications can get system information through/PROC, and can change certain parameters of the kernel. Because the information of the system, such as the process, is dynamically changed, so when a user or application reads a file from the/proc dire

Tags: code proc rate BLE 1.0 double can return pen#define SYS_MEM_NAME_LEN 20#define SYS_MEM_BUFF_LEN 256#define SYS_100_PERSENT 100intget_sysMemUsage(){ FILE *fd; char buff[SYS_MEM_BUFF_LEN]; double mem_used_rate; char name01[SYS_MEM_NAME_LEN]; unsigned long mem_total; char name02[SYS_MEM_NAME_LEN]; char name11[SYS_MEM_NAME_LEN]; unsigned long mem_free; char name12[SYS_MEM_NAME_LEN]; fd = fopen ("/proc/meminfo", "r"); fgets (buff, sizeof(buff), fd); sscanf (buff, "%s %lu %s\n", name01, mem_tota

In the last blog post, we talked about how to get the results of executing the command line in Linux, and eventually get CPU usage, but unfortunately it is:1) via Top-n 1 | The CPU usage obtained by the GREP CPU is not refreshed, and the results of the first execution are only available for each read.2) may cause a conflict on signalThe new code solves the principle:Calculate CPU usage by reading data from/proc/stat#include string.h> #define FILEBUFF

Calculate Linux system memory utilization correctlyFor the operating system, buffers and cached are already used (Mem: this line)memfree=total-used314952=24946552-24631600For the application (corresponding to the-/+ Buffers/cache line) memfree=buffers+cached+free19536392=152116+19069324+314952 [[emailprotected] bin] # free total used free shared buffers mem:37 820 37316 504 0 704 swap:0 0 0total:37

Linux Shell View disk partition, memory usage, CPU utilization#!/bin/bash#disk_used_ratelocation=/dev/xvdbdisk_used_rate=$ (df-h | grep $Location | awk ' {print $} ') echo "disk used Ra TE: $Disk _used_rate "#memory_used_rateLoadMemory =$ (cat/proc/meminfo | awk ' {print $} ') total=$ (echo $LoadMemory | awk ' { Print $ ') free1=$ (echo $LoadMemory | awk ' {print $} ') free2=$ (echo $LoadMemory | awk ' {pri

Today ran a very time-consuming bulk insert operation. When using the top command to view the CPU and memory usage, the CPU was checked by 120%. Never noticed before. By pressing 1 on the big keyboard in the case of top, the number of cores viewed by the CPU is 4 cores.By looking through the Web, the top command shows the total number of CPUs your program consumes, which means that if you are a 4-core CPU, the maximum CPU occupancy is up to 400%,top and all usage rates are added together.Here we

1, the network of O Internet model 650) this.width=650; "title=" image "style=" border-left-0px; border-right-width:0px; Background-image:none; border-bottom-width:0px; padding-top:0px; padding-left:0px; padding-right:0px; border-top-width:0px "border=" 0 "alt=" image "src=" http://s3.51cto.com/wyfs02/M01/73/6C/wKiom1X9MT-xfkE_ Aaderuzjjyy990.jpg "width=" 385 "height=" 209 "/> Physical layer: Electrical signals, n