我试图使用时间()来测量我的程序的各个点。
我不明白的是为什么前后的值是一样的?我知道这不是分析我的程序的最佳方式,我只是想看看需要多长时间。
printf("**MyProgram::before time= %ld\n", time(NULL));
doSomthing();
doSomthingLong();
printf("**MyProgram::after time= %ld\n", time(NULL));
我试过:
struct timeval diff, startTV, endTV;
gettimeofday(&startTV, NULL);
doSomething();
doSomethingLong();
gettimeofday(&endTV, NULL);
timersub(&endTV, &startTV, &diff);
printf("**time taken = %ld %ld\n", diff.tv_sec, diff.tv_usec);
我如何读取**时间花费= 0 26339的结果?这是否意味着26339纳秒= 26.3毫秒?
**时间= 4 45025,这是否意味着4秒25毫秒?
c++ std::chrono具有跨平台的明显优势。
然而,与POSIX clock_gettime()相比,它也引入了显著的开销。
在我的Linux机器上,所有std::chrono::xxx_clock::now()味道的执行大致相同:
std::chrono::system_clock::now()
std::chrono::steady_clock::now()
std::chrono::high_resolution_clock::now()
虽然POSIX clock_gettime(CLOCK_MONOTONIC, &time)应该与steady_clock::now()相同,但它要快x3倍以上!
这是我的测试,为了完整性。
#include <stdio.h>
#include <chrono>
#include <ctime>
void print_timediff(const char* prefix, const struct timespec& start, const
struct timespec& end)
{
double milliseconds = end.tv_nsec >= start.tv_nsec
? (end.tv_nsec - start.tv_nsec) / 1e6 + (end.tv_sec - start.tv_sec) * 1e3
: (start.tv_nsec - end.tv_nsec) / 1e6 + (end.tv_sec - start.tv_sec - 1) * 1e3;
printf("%s: %lf milliseconds\n", prefix, milliseconds);
}
int main()
{
int i, n = 1000000;
struct timespec start, end;
// Test stopwatch
clock_gettime(CLOCK_MONOTONIC, &start);
for (i = 0; i < n; ++i) {
struct timespec dummy;
clock_gettime(CLOCK_MONOTONIC, &dummy);
}
clock_gettime(CLOCK_MONOTONIC, &end);
print_timediff("clock_gettime", start, end);
// Test chrono system_clock
clock_gettime(CLOCK_MONOTONIC, &start);
for (i = 0; i < n; ++i)
auto dummy = std::chrono::system_clock::now();
clock_gettime(CLOCK_MONOTONIC, &end);
print_timediff("chrono::system_clock::now", start, end);
// Test chrono steady_clock
clock_gettime(CLOCK_MONOTONIC, &start);
for (i = 0; i < n; ++i)
auto dummy = std::chrono::steady_clock::now();
clock_gettime(CLOCK_MONOTONIC, &end);
print_timediff("chrono::steady_clock::now", start, end);
// Test chrono high_resolution_clock
clock_gettime(CLOCK_MONOTONIC, &start);
for (i = 0; i < n; ++i)
auto dummy = std::chrono::high_resolution_clock::now();
clock_gettime(CLOCK_MONOTONIC, &end);
print_timediff("chrono::high_resolution_clock::now", start, end);
return 0;
}
这是我用gcc7.2 -O3编译时得到的输出:
clock_gettime: 24.484926 milliseconds
chrono::system_clock::now: 85.142108 milliseconds
chrono::steady_clock::now: 87.295347 milliseconds
chrono::high_resolution_clock::now: 84.437838 milliseconds
下面是一个简单的类,它将在指定的持续时间单位内打印它进入和离开作用域之间的持续时间:
#include <chrono>
#include <iostream>
template <typename T>
class Benchmark
{
public:
Benchmark(std::string name) : start(std::chrono::steady_clock::now()), name(name) {}
~Benchmark()
{
auto end = std::chrono::steady_clock::now();
T duration = std::chrono::duration_cast<T>(end - start);
std::cout << "Bench \"" << name << "\" took: " << duration.count() << " units" << std::endl;
}
private:
std::string name;
std::chrono::time_point<std::chrono::steady_clock> start;
};
int main()
{
Benchmark<std::chrono::nanoseconds> bench("for loop");
for(int i = 0; i < 1001000; i++){}
}
使用示例:
int main()
{
Benchmark<std::chrono::nanoseconds> bench("for loop");
for(int i = 0; i < 100000; i++){}
}
输出:
Bench "for loop" took: 230656 units