C语言实现任意时间戳和时间的相互转换

C语言实现任意时间戳和时间的相互转换

1. 代码如下：

#include <stdio.h>
#include <time.h>
#include <sys/time.h>
#include <unistd.h>

#define yyyy 0
#define MM 1
#define dd 2
#define HH 3
#define mm 4
#define ss 5

const char dayOfMonth[] = {31,28,31,30,31,30,31,31,30,31,30,31};

/*
* ts: 秒数时间戳
* min：时区偏移分钟数，比如+8时区，则min=60*8=480，+9:30时区，则min=9*60+30=570
* buffer ： 返回计算得到的年月日时分秒缓冲区
*/
void praseTimestamp(long ts,int min,int *buffer) 
{
    //闰年标记
    char leapYear = 0;
    int i = 0;
	int remain = 0;
	int mon_day = 0;
	int ts_yu = 0;
    
    //计算时间戳的天数
    long days = (ts + 60 * min) / 86400;
	ts_yu = (ts + 60 * min) % 86400;
	if(ts_yu >= 0){
		days += 1;
	}
	for(i = 1970,remain = days; remain > 0; i++)
	{
		if(((i % 4 == 0) && (i % 100 != 0)) || (i % 400 == 0)){
			remain -= 366;
			leapYear = 1;
		}else{
			remain -= 365;
			leapYear = 0;
		}
	}
	remain = leapYear?(remain+366):(remain+365);
	buffer[yyyy] = i-1;
    
    //剩下一年内的天数 用12个月天数循环减
    //直到剩下天数<=0 则找到对应月份
    //剩下天数则为日期
    for(i=0;i<12;i++){
        //闰年2月补一天
        mon_day = (i == 1 && leapYear) ? (dayOfMonth[i] + 1) : dayOfMonth[i];
        
        //days复用于暂存减去当前月份天数的剩余天数
        days = remain - mon_day;
        if(days <= 0){
            
            buffer[MM] = i + 1;
            if(days == 0){
                buffer[dd] = mon_day;
            }else{
                    buffer[dd] = remain;
            }
            break;
        }
        remain = (int)days;
    }
    
    buffer[ss] = (int)(ts % 60);

    buffer[mm] = (int)((ts + min*60) /60 % 60);
	
	buffer[HH] = (int)(ts_yu/3600);
}

/*
* min：时区偏移分钟数，比如+8时区，则min=60*8=480，+9:30时区，则min=9*60+30=570
* buffer：存放年月日时分秒的数组，比如buf[]={2038,5,5,12,30,30};
* return : 返回计算得到的时间戳
*/
long long makeTimestamp(int min,int *buffer) 
{
    long long ts;
    char leapYear = 0;
    int i = 0;
	int days = 0;

    //确定当前年是否闰年
    if(((buffer[yyyy] % 4 == 0) && (buffer[yyyy] % 100 != 0))
		|| (buffer[yyyy] % 400 == 0)){
		leapYear = 1;
	}
	//年份对应1970-1-1偏移的天数
    for(i = 1970; i < buffer[yyyy]; i++){
		if(((i % 4 == 0) && (i % 100 != 0)) || (i % 400 == 0)){
			days += 366;
		}else{
			days += 365;
		}
	}

    ts =  days * 86400L;
    
    //处理当前月之前月份天数的秒数
    for(i=0;i<buffer[MM]-1;i++){
        if(i==1 && leapYear){
            ts += (dayOfMonth[i] + 1) * 86400L;
        }else{
            ts += dayOfMonth[i] * 86400L;
        }
    }
    
    //处理当前日期之前的天数的秒数
    ts += 86400L * (buffer[dd] - 1);
    
    //处理当前天小时的秒数
    ts += 3600L * (buffer[HH]);
    
    //处理当前天分钟的秒数
    ts += 60L * (buffer[mm]);
    
    //加上当前天的秒数
    ts += buffer[ss];
    
    //减去时区差
    ts -= (min * 60);
    
    return ts;
}

int main(char* argv[], int argc)
{
	int buf[7] = {2200,12,31,23,59,59};
	
	int buf3[10] = {0};
	int time_zone_min = 0;
	long long int time_start = 3599;
	long long int time_check = 0;
	
	/* 计算+8时区的2200-12-31 23:59:59对应的时间戳 */
	printf("2200-12-31 23:59:59----->zone:+8时区 %lld\n",makeTimestamp(480, buf));
	/* 计算+3时区的4115422800对应的年月日时分秒 */
	praseTimestamp(4115422800, 180, buf3);
	printf("4115422800---->zone:+3时区 %d/%d/%d %d:%d:%d\n", buf3[0], buf3[1], buf3[2], buf3[3], buf3[4], buf3[5]);
	/* 从-12时区到+12时区，每隔30分钟一个时区，遍历计算 */
	for(time_zone_min = -720; time_zone_min <= 720; time_zone_min+=30){
		printf("time_zone_min = %d start\n", time_zone_min);
		for(time_start = 75599; time_start <= 7289625599; time_start += 360){
			praseTimestamp(time_start, time_zone_min, buf3);
			time_check = makeTimestamp(time_zone_min, buf3);
			/* 若转换有异常则打印 */
			if(time_check != time_start){
				printf("error：time_check:%lld,time_start:%lld, time_zone_min:%d,%d/%d/%d-%02d:%02d:%02d\n",time_check,time_start, time_zone_min, buf3[0], buf3[1], buf3[2], buf3[3], buf3[4], buf3[5]); 
			}
		}
	}
	
	return 0;
}

2. 运行打印如下：
   2200-12-31 23:59:59----->zone:+8时区 7289625599
   4115422800---->zone:+3时区 2100/5/31 8:0:0
   time_zone_min = -720 start
   time_zone_min = -690 start
   time_zone_min = -660 start
   time_zone_min = -630 start
   …
   …
   time_zone_min = 690 start
   time_zone_min = 720 start

参考: https://github.com/Yanye0xFF/Timestamp

No pains, no gains.