两个日期之间的天数C ++ [英] Number of days between two dates C++

问题描述

我看到C＃，Java的例子，但对于C ++，我无法找到解决方案来计算两个日期之间的天数。

例如2012-01-24和2013-01-08

谢谢！

解决方案

一种方式。

  #include< iostream> 
 #include< ctime> 
 
 int main（）
 {
 struct std :: tm a = {0,0,0,24,5,104}; / * 2004年6月24日* / 
 struct std :: tm b = {0,0,0,5,6,104}; / * 2004年7月5日* / 
 std :: time_t x = std :: mktime（& a）; 
 std :: time_t y = std :: mktime（& b）; 
 if（x！=（std :: time_t）（ -  1）&& y！=（std :: time_t）（ -  1））
 {
 double difference = std :: difftime（y，x）/（60 * 60 * 24）; 
 std :: cout<<< std :: ctime（& x）; 
 std :: cout<<< std :: ctime（& y）; 
 std :: cout<<< difference =<<差异< 天< std :: endl; 
} 
 return 0; 
} 
  

我的输出

 星期四24 01:00:00 2004 
 Mon Jul 05 01:00:00 2004 
差= 11天
  

以下是对原作者帖子的评论

I saw examples for C#, Java, but for C++ i cant find solution to calculate how many days between two dates.

For example between 2012-01-24 and 2013-01-08

Thanks!

解决方案

This is one way.

#include <iostream>
#include <ctime>

int main()
{
    struct std::tm a = {0,0,0,24,5,104}; /* June 24, 2004 */
    struct std::tm b = {0,0,0,5,6,104}; /* July 5, 2004 */
    std::time_t x = std::mktime(&a);
    std::time_t y = std::mktime(&b);
    if ( x != (std::time_t)(-1) && y != (std::time_t)(-1) )
    {
        double difference = std::difftime(y, x) / (60 * 60 * 24);
        std::cout << std::ctime(&x);
        std::cout << std::ctime(&y);
        std::cout << "difference = " << difference << " days" << std::endl;
    }
    return 0;
}

my output

Thu Jun 24 01:00:00 2004
Mon Jul 05 01:00:00 2004
difference = 11 days

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