时间不匹配

Question

struct timeval start, end, duration;
gettimeofday(&start, NULL);
res = curl_easy_perform(curl);
gettimeofday(&end, NULL);
timersub(&end, &start, &duration);
tm* startTime = localtime(&start.tv_sec);
tm* endTime = localtime(&end.tv_sec);
char buf[64];
strftime(buf, sizeof(buf), "%Y-%m-%d %H:%M:%S", startTime);
char buf2[64];
strftime(buf2, sizeof(buf2), "%Y-%m-%d %H:%M:%S", endTime);
ofstream timeFile;
timeFile.open ("timingSheet.txt");
timeFile << fixed << showpoint;
timeFile << setprecision(6);
timeFile << "Duration: " << duration.tv_sec << "." << duration.tv_usec << " seconds n";
timeFile << "Start time: " <<  buf <<"." << start.tv_usec << "n";
timeFile << "End time: " <<  buf2 <<"." << end.tv_usec << "n";
timeFile.close();

当我运行此代码时，我得到以下输出：

Duration: 3.462243 seconds 
Start time: 2012-05-15 17:14:07.432613
End time: 2012-05-15 17:14:07.894856

令我困惑的是持续时间值与开始和结束时间不匹配。这两个日期仅相差几微秒。这有什么原因吗？

谢谢！

Answer 1

localtime返回一个静态分配的缓冲区，你调用它两次，所以StartTime和EndTime是相同的。每次调用后，您需要直接将其复制到另一个缓冲区中。

tm* startTime = localtime(&start.tv_sec); 
char buf[64]; 
strftime(buf, sizeof(buf), "%Y-%m-%d %H:%M:%S", startTime); 
tm* endTime = localtime(&end.tv_sec); 
char buf2[64]; 
strftime(buf2, sizeof(buf2), "%Y-%m-%d %H:%M:%S", endTime); 

编辑：你也可以这样写：

tm* pTimeBuf = localtime(&start.tv_sec); 
char buf[64]; 
strftime(buf, sizeof(buf), "%Y-%m-%d %H:%M:%S", pTimeBuf); 
localtime(&end.tv_sec); // NB. I don't store th return value (since I have it already)
char buf2[64]; 
strftime(buf2, sizeof(buf2), "%Y-%m-%d %H:%M:%S", pTimeBuf); 

Answer 2

我在这里同意 Edwin 的观点，只是一个小的修改，最好使用 localtime_r 线程安全版本而不是本地时间

struct tm startTime,endTime;
memset(&startTime,0,sizeof(struct tm)); //Advisable but not necessary
memset(&endTime,0,sizeof(struct tm)); //Advisable but not necessary
localtime_r(&start.tv_sec, &startTime);
localtime_r(&end.tv_sec, &endTime);