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  • 时间处理:计算下一天日期,如输入"2004/12/31"(注释2014年12月31日),则输出"2005/1/1".

    /*
     ============================================================================
     Name         : Exercise.c
     Author       : haier
     Version      : 0.01
     Copyright   : Your copyright notice
     Description : Ansi-style, Compile by Code:Blocks, Platform Linux
     ============================================================================
     */
    
    /*
    思路:首先求出输入时间到1970-1-1总秒数,然后加上一天时间后转换成日历时间并输出。
    */
    
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <time.h>
    #include <ctype.h>
    
    #define SECONDS_PER_DAY (60*60*24)
    
    struct date
    {
        unsigned year;
        unsigned month;
        unsigned day;
    } myDate;
    
    void nextDay(struct date myDate)
    {
        struct tm time,*tm_ptr;
        time_t seconds;
    
        /*填充struct tm*/
        time.tm_year = myDate.year - 1900;
        time.tm_mon  = myDate.month - 1;
        time.tm_mday = myDate.day;
        time.tm_hour = 12; //默认为该日12:00:00
        time.tm_min  = 0;
        time.tm_sec  = 0;
        seconds=mktime(&time)+SECONDS_PER_DAY;  //转换tm结构为time_t类型值,并加一天
    
        /*转换并打印日期*/
        tm_ptr = gmtime(&seconds);
        printf("%d/%d/%d
    ",tm_ptr->tm_year+1900, tm_ptr->tm_mon+1, tm_ptr->tm_mday);
    }
    
    int main()
    {
        char strTime[20];
        char *p;
        int inputTimes=3;
    
        /*输入处理,允许输入3次*/
        do
        {
            inputTimes--;
            printf("Please input the date(YYYY/MM/DD) : ");
            scanf("%s",strTime);
    
            /*转换时间字符串*/
            p = strtok(strTime, "/");
            myDate.year=atoi(p);
            p = strtok(NULL, "/");
            myDate.month=atoi(p);
            p = strtok(NULL, "/");
            myDate.day=atoi(p);
    
            if(myDate.year<1970 || myDate.month<=0 || myDate.month>12 || myDate.day<=0 || myDate.day >31)
            {
                printf("error in input, please try again !
    ");
            }
            else
            {
                 break;
            }
    
        }while(inputTimes>0);
    
        if(inputTimes>0)
        {    
            nextDay(myDate);
        }
    
        return 0;
    }

    运行示例:

    image

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  • 原文地址:https://www.cnblogs.com/274914765qq/p/4331742.html
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