我正在尝试在我的程序中使用“usage”统计信息来获取类似于 thời gian 的数据工具。但是,我很确定我做错了什么。这些值似乎是正确的,但有时可能有点奇怪。我没有在网上找到好的资源。有人知道如何做得更好吗?
抱歉代码太长。
class StopWatch {
public:
void start() {
getrusage(RUSAGE_SELF, &m_begin);
gettimeofday(&m_tmbegin, 0);
}
void stop() {
getrusage(RUSAGE_SELF, &m_end);
gettimeofday(&m_tmend, 0);
timeval_sub(m_end.ru_utime, m_begin.ru_utime, m_diff.ru_utime);
timeval_sub(m_end.ru_stime, m_begin.ru_stime, m_diff.ru_stime);
timeval_sub(m_tmend, m_tmbegin, m_tmdiff);
}
void printf(std::ostream& out) const {
using namespace std;
timeval const& utime = m_diff.ru_utime;
timeval const& stime = m_diff.ru_stime;
format_time(out, utime);
out << "u ";
format_time(out, stime);
out << "s ";
format_time(out, m_tmdiff);
}
private:
rusage m_begin;
rusage m_end;
rusage m_diff;
timeval m_tmbegin;
timeval m_tmend;
timeval m_tmdiff;
static void timeval_add(timeval const& a, timeval const& b, timeval& ret) {
ret.tv_usec = a.tv_usec + b.tv_usec;
ret.tv_sec = a.tv_sec + b.tv_sec;
if (ret.tv_usec > 999999) {
ret.tv_usec -= 1000000;
++ret.tv_sec;
}
}
static void timeval_sub(timeval const& a, timeval const& b, timeval& ret) {
ret.tv_usec = a.tv_usec - b.tv_usec;
ret.tv_sec = a.tv_sec - b.tv_sec;
if (a.tv_usec < b.tv_usec) {
ret.tv_usec += 1000000;
--ret.tv_sec;
}
}
static void format_time(std::ostream& out, timeval const& tv) {
using namespace std;
long usec = tv.tv_usec;
while (usec >= 1000)
usec /= 10;
out << tv.tv_sec << '.' << setw(3) << setfill('0') << usec;
}
}; // class StopWatch
目的是什么:
while (usec >= 1000)
usec /= 10;
我了解到您需要 usec 的最高三位数字;在这种情况下,我能想到的最直接的方法是将 usec 除以 1000,然后用它完成。
测试用例:
- 999999 ⇒ 999
- 99999 ⇒ 999(应该是099)
- 9999 ⇒ 999(应该是009)
- 999 ⇒ 999(应该是000)
Tôi là một lập trình viên xuất sắc, rất giỏi!