所以我现在已经解决了这个问题一段时间了,在成员的帮助下,我差不多完成了.我的上一个问题是上面的问题.
如果需要,我需要将cout时间格式化为01:01:01,但是,我的输出是1:1:1
问题出在这里:
std::ostream& operator<< (ostream& os, const MyTime& m)
{
os << setfill('0') << m.hours << ":" << setfill ('0') << m.minutes << ":" << setfill ('0') << m.seconds;
return os;
}
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这是整个代码.
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <iomanip>
using namespace std;
struct MyTime { int hours, minutes, seconds; };
MyTime DetermineElapsedTime(const MyTime *t1, const MyTime *t2);
const int hourSeconds = 3600;
const int minSeconds = 60;
const int dayHours = 24;
const char zero = 0;
MyTime DetermineElapsedTime(const MyTime *t1, const MyTime *t2)
{
long hourDiff = ((t2->hours * hourSeconds) - (t1->hours * hourSeconds));
int timeHour = hourDiff / hourSeconds;
long minDiff = ((t2->minutes * minSeconds) - (t1->minutes * minSeconds));
int timeMin = minDiff / minSeconds;
int timeSec = (t2->seconds - t1 -> seconds);
MyTime time;
time.hours = timeHour;
time.minutes = timeMin;
time.seconds = timeSec;
return time;
}
std::ostream& operator<< (ostream& os, const MyTime& m)
{
os << setfill('0') << m.hours << ":" << setfill ('0') << m.minutes << ":" << setfill ('0') << m.seconds;
return os;
}
int main(void)
{
char delim1, delim2;
MyTime tm, tm2;
cout << "Input two formats for the time. Separate each with a space. Ex: hr:min:sec\n";
cin >> tm.hours >> delim1 >> tm.minutes >> delim2 >> tm.seconds;
cin >> tm2.hours >> delim1 >> tm2.minutes >> delim2 >> tm2.seconds;
if (tm2.hours <= tm.hours && tm2.minutes <= tm.minutes && tm2.seconds <= tm.seconds)
{
tm2.hours += dayHours;
}
cout << DetermineElapsedTime(&tm, &tm2);
return 0;
}
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任何解决这个问题的见解将不胜感激.如果您需要更多信息,请询问.
你还需要调用setw让流知道它必须填充输出.请记住,与之相反setfill,setw仅适用于下一个输出.
std::ostream& operator<< (ostream& os, const MyTime& m)
{
return os << setfill('0') << setw(2) << m.hours << ":"
<< setw(2) << m.minutes << ":"
<< setw(2) << m.seconds;
}
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