如何获得当前的系统时间和转换在数组

Question

我想分别获得时间和日期，并将其放入数组中。我希望调用函数是int date(int Day,int Month,int Year)，但这是不正确的。如何定义函数的年、月、日、时、分、秒参数来使用它们?

using namespace std;
int date() {
    time_t currentTime;
    struct tm *localTime;
    time(&currentTime);                  
    localTime = localtime(&currentTime);
     int Day = localTime->tm_mday;
     int Month = localTime->tm_mon + 1;
     int Year = localTime->tm_year + 1900;
    return (0);
}
int time() {
    time_t currentTime;
    struct tm *localTime;
    time(&currentTime);                   
    localTime = localtime(&currentTime); 
    int Hour = localTime->tm_hour;
    int Min = localTime->tm_min;
    int Sec = localTime->tm_sec;
    return (0);
}
int main() {
    unsigned int new_date=date();
    char write[4];
    memcpy(write,&new_date,4);
    unsigned int new_time=time();
       char wrt[4];
       memcpy(wrt,&new_time,4);

    }

Answer 1

必须声明函数通过引用传递参数:

              // pass arguments by reference to change them in the function 
int date(int &Day, int &Month, int&Year) {
    time_t currentTime;
    struct tm *localTime;
    time(&currentTime);                  
    localTime = localtime(&currentTime);
     Day = localTime->tm_mday;   // use the reference argument
     Month = localTime->tm_mon + 1;
     Year = localTime->tm_year + 1900;
    return (0);
}

您可以在main()中这样使用它:

int d,m,y; 
date(d,m,y); 
cout << d<<"/"<<m<<"/"<<y<<endl;

你当然可以对时间做同样的事情。