如何在 Linux 机器上记录应用程序的性能？我不会有 IDE。
理想情况下，我需要一个将附加到进程并记录定期快照的应用程序:
- 内存使用情况
- 线程数
- CPU 使用率
最佳答案
Ideally, I need an application that will attach to a process and log periodic snapshots of:
- memory usage
- number of threads
- CPU usage
好吧，为了收集有关您的进程的此类信息，您实际上并不需要 Linux 上的分析器。
您可以在批处理模式下使用
top。它以批处理模式运行，直到它被杀死或完成 N 次迭代:top -b -p `pidof a.out`或
top -b -p `pidof a.out` -n 100你会得到这个:
$ top -b -p `pidof a.out` top - 10:31:50 up 12 days, 19:08, 5 users, load average: 0.02, 0.01, 0.02 Tasks: 1 total, 0 running, 1 sleeping, 0 stopped, 0 zombie Cpu(s): 0.0%us, 0.0%sy, 0.0%ni,100.0%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 16330584k total, 2335024k used, 13995560k free, 241348k buffers Swap: 4194296k total, 0k used, 4194296k free, 1631880k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 24402 SK 20 0 98.7m 1056 860 S 43.9 0.0 0:11.87 a.out top - 10:31:53 up 12 days, 19:08, 5 users, load average: 0.02, 0.01, 0.02 Tasks: 1 total, 0 running, 1 sleeping, 0 stopped, 0 zombie Cpu(s): 0.9%us, 3.7%sy, 0.0%ni, 95.5%id, 0.0%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 16330584k total, 2335148k used, 13995436k free, 241348k buffers Swap: 4194296k total, 0k used, 4194296k free, 1631880k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 24402 SK 20 0 98.7m 1072 860 S 19.0 0.0 0:12.44 a.out您可以使用
ps(例如在 shell 脚本中)ps --format pid,pcpu,cputime,etime,size,vsz,cmd -p `pidof a.out`
I need some means of recording the performance of an application on a Linux machine
如果您的 Linux 内核大于 2.6.32 或 OProfile，您需要使用
perf来执行此操作如果它更旧。这两个程序都不需要您来检测您的程序(如 Gprof 需要)。但是，为了在perf中正确获取调用图，您需要使用 -fno-omit-frame-pointer 构建程序。例如:g++ -fno-omit-frame-pointer -O2 main.cpp.
至于Linux perf:
记录性能数据:
perf record -p `pidof a.out`或录制 10 秒:
perf record -p `pidof a.out` sleep 10或者用调用图记录()
perf record -g -p `pidof a.out`分析记录的数据
perf report --stdio perf report --stdio --sort=dso -g none perf report --stdio -g none perf report --stdio -g开启 RHEL 6.3 允许读取/boot/System.map-2.6.32-279.el6.x86_64，所以我一般在做的时候加上--kallsyms=/boot/System.map-2.6.32-279.el6.x86_64业绩报告:
perf report --stdio -g --kallsyms=/boot/System.map-2.6.32-279.el6.x86_64
这里我写了一些关于使用 Linux `perf` 的更多信息:首先 - 这是tutorial about Linux profiling with perf
如果您的 Linux 内核大于 2.6.32，您可以使用 perf，如果它较旧，您可以使用 OProfile。这两个程序都不需要你来检测你的程序(就像 Gprof 需要的那样)。但是，为了在 perf 中正确获取调用图，您需要使用
-fno-omit-frame-pointer构建程序。例如:g++ -fno-omit-frame-pointer -O2 main.cpp.您可以使用 perf top 查看对您的应用程序的“实时”分析:
sudo perf top -p `pidof a.out` -K
或者您可以记录正在运行的应用程序的性能数据并在之后进行分析:
记录性能数据:
perf record -p `pidof a.out`或录制 10 秒:
perf record -p `pidof a.out` sleep 10或者用调用图记录()
perf record -g -p `pidof a.out`分析记录的数据
perf report --stdio
perf report --stdio --sort=dso -g none
perf report --stdio -g none
perf report --stdio -g
或者您可以记录应用程序的性能数据并在之后分析它们，只需以这种方式启动应用程序并等待它退出:
perf record ./a.out
这是一个分析测试程序的示例。
测试程序在文件main.cpp中(main.cpp在答案的底部):
我是这样编译的:
g++ -m64 -fno-omit-frame-pointer -g main.cpp -L. -ltcmalloc_minimal -o my_test
我使用 libmalloc_minimial.so，因为它是用 -fno-omit-frame-pointer 编译的，而 libc malloc 似乎是在没有这个选项的情况下编译的。然后我运行我的测试程序:
./my_test 100000000
然后我记录一个正在运行的进程的性能数据:
perf record -g -p `pidof my_test` -o ./my_test.perf.data sleep 30
然后我分析每个模块的负载:
perf report --stdio -g none --sort comm,dso -i ./my_test.perf.data
# Overhead Command Shared Object
# ........ ....... ............................
#
70.06% my_test my_test
28.33% my_test libtcmalloc_minimal.so.0.1.0
1.61% my_test [kernel.kallsyms]
然后分析每个函数的负载:
perf report --stdio -g none -i ./my_test.perf.data | c++filt
# Overhead Command Shared Object Symbol
# ........ ....... ............................ ...........................
#
29.30% my_test my_test [.] f2(long)
29.14% my_test my_test [.] f1(long)
15.17% my_test libtcmalloc_minimal.so.0.1.0 [.] operator new(unsigned long)
13.16% my_test libtcmalloc_minimal.so.0.1.0 [.] operator delete(void*)
9.44% my_test my_test [.] process_request(long)
1.01% my_test my_test [.] operator delete(void*)@plt
0.97% my_test my_test [.] operator new(unsigned long)@plt
0.20% my_test my_test [.] main
0.19% my_test [kernel.kallsyms] [k] apic_timer_interrupt
0.16% my_test [kernel.kallsyms] [k] _spin_lock
0.13% my_test [kernel.kallsyms] [k] native_write_msr_safe
and so on ...
然后分析调用链:
perf report --stdio -g graph -i ./my_test.perf.data | c++filt
# Overhead Command Shared Object Symbol
# ........ ....... ............................ ...........................
#
29.30% my_test my_test [.] f2(long)
|
--- f2(long)
|
--29.01%-- process_request(long)
main
__libc_start_main
29.14% my_test my_test [.] f1(long)
|
--- f1(long)
|
|--15.05%-- process_request(long)
| main
| __libc_start_main
|
--13.79%-- f2(long)
process_request(long)
main
__libc_start_main
15.17% my_test libtcmalloc_minimal.so.0.1.0 [.] operator new(unsigned long)
|
--- operator new(unsigned long)
|
|--11.44%-- f1(long)
| |
| |--5.75%-- process_request(long)
| | main
| | __libc_start_main
| |
| --5.69%-- f2(long)
| process_request(long)
| main
| __libc_start_main
|
--3.01%-- process_request(long)
main
__libc_start_main
13.16% my_test libtcmalloc_minimal.so.0.1.0 [.] operator delete(void*)
|
--- operator delete(void*)
|
|--9.13%-- f1(long)
| |
| |--4.63%-- f2(long)
| | process_request(long)
| | main
| | __libc_start_main
| |
| --4.51%-- process_request(long)
| main
| __libc_start_main
|
|--3.05%-- process_request(long)
| main
| __libc_start_main
|
--0.80%-- f2(long)
process_request(long)
main
__libc_start_main
9.44% my_test my_test [.] process_request(long)
|
--- process_request(long)
|
--9.39%-- main
__libc_start_main
1.01% my_test my_test [.] operator delete(void*)@plt
|
--- operator delete(void*)@plt
0.97% my_test my_test [.] operator new(unsigned long)@plt
|
--- operator new(unsigned long)@plt
0.20% my_test my_test [.] main
0.19% my_test [kernel.kallsyms] [k] apic_timer_interrupt
0.16% my_test [kernel.kallsyms] [k] _spin_lock
and so on ...
所以此时您知道您的程序将时间花在了哪里。
这是用于测试的 main.cpp 文件:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
time_t f1(time_t time_value)
{
for (int j = 0; j < 10; ++j) {
++time_value;
if (j%5 == 0) {
double *p = new double;
delete p;
}
}
return time_value;
}
time_t f2(time_t time_value)
{
for (int j = 0; j < 40; ++j) {
++time_value;
}
time_value = f1(time_value);
return time_value;
}
time_t process_request(time_t time_value)
{
for (int j = 0; j < 10; ++j) {
int *p = new int;
delete p;
for (int m = 0; m < 10; ++m) {
++time_value;
}
}
for (int i = 0; i < 10; ++i) {
time_value = f1(time_value);
time_value = f2(time_value);
}
return time_value;
}
int main(int argc, char* argv2[])
{
int number_loops = argc > 1 ? atoi(argv2[1]) : 1;
time_t time_value = time(0);
printf("number loops %d\n", number_loops);
printf("time_value: %d\n", time_value);
for (int i = 0; i < number_loops; ++i) {
time_value = process_request(time_value);
}
printf("time_value: %ld\n", time_value);
return 0;
}
关于Linux 应用程序分析，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/2229336/