C语言中如何计算时间差

Question

如何将两个包含“年月日时分”的5个字节数据进行比较，最后得出一个时间间隔（总分钟数）？
其中的年份用后两位表示，如2010年8月31日9时26分表示为：1008310926
最后输出必须是时间差（总的分钟）
补充：在KEIL C中编译没有time.h这个头文件调用怎么办？

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void main()
{
unsigned char time1[] = { 10, 8, 31, 9, 26 };
unsigned char time2[] = { 10, 8, 31, 9, 50 };

struct tm t1 = {0};
struct tm t2 = {0};

time_t _t1;
time_t _t2;

double diff;

t1.tm_year = time1[0] + 100;
t1.tm_mon = time1[1];
t1.tm_mday = time1[2];
t1.tm_hour = time1[3];
t1.tm_min = time1[4];

t2.tm_year = time2[0] + 100;
t2.tm_mon = time2[1];
t2.tm_mday = time2[2];
t2.tm_hour = time2[3];
t2.tm_min = time2[4];

_t1 = _mkgmtime( &t1 );
_t2 = _mkgmtime( &t2 );

diff = difftime(_t2, _t1 );

printf( "相差 %.0f 分钟\n", diff / 60 );

}本回答被提问者采纳