由于我在这里看不到任何关于perf在Linux上对内核和用户应用程序进行性能分析的相对较新的工具，因此我决定添加此信息。
首先-这是关于Linux性能分析的教程perf
perf如果您的Linux内核大于2.6.32或oprofile更旧，则可以使用。这两个程序都不需要您来检测您的程序（就像require一样gprof）。但是，为了正确获取调用图，perf您需要使用构建程序-fno-omit-frame-pointer。例如：g++ -fno-omit-frame-pointer -O2 main.cpp。
您可以通过以下方式对应用程序进行“实时”分析perf top：
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，然后对其进行分析：
1）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
或使用通话记录（）进行记录
perf record -g -p `pidof a.out`
2）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，然后通过以这种方式启动应用程序并等待其退出来对其进行分析：
perf record ./a.out
这是分析测试程序的示例
测试程序位于文件main.cpp中（我将main.cpp放在消息的底部）：
我以这种方式编译它：
g++ -m64 -fno-omit-frame-pointer -g main.cpp -L. -ltcmalloc_minimal -o my_test
我使用libmalloc_minimial.so它是因为-fno-omit-frame-pointerlibc malloc似乎是在没有此选项的情况下编译的。然后我运行测试程序
./my_test 100000000
然后，我记录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：
性能报告--stdio -g无--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：
性能报告--stdio -g none -i ./my_test.perf.data | C ++过滤
# 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 ...
然后分析呼叫链：
性能报告--stdio -g图-i ./my_test.perf.data | C ++过滤
# 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 ...
这样，您就知道了程序在哪里花费时间。
这是用于测试的main.cpp：
#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;
}