c++计算不同日期间隔天数

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有时候需要计算2个日期间隔天数,在python,vb,c#等内置了DateTime结构的语言中非常方便,有些提供了DateDiff函数,有些可以直接作差.c++的boost库也有类似实现,但是如果需要自己写代码实现,如何解决呢?

答案是使用Julian Day Number(儒略日),具体参考这里.将公元纪年法转为儒略日数字(float)后直接做差即可实现.c++实现代码如下(公元纪年法或称格里高利纪年法与儒略日互相转换):

#include <iostream>
using namespace std;
 
// convert Gregorian calendar date to Julian Day Number
float getJulianDay(int year, int month, int day)
{
    int a = (14-month)/12;
    int y = year+4800-a;
    int m = month+12*a-3;
    return (float)day+(float)(153*m+2)/5+365*y+(float)y/4-32083;
}
 
// convert Julian Day Number to Gregorian date
void GetGregorianDay(float jd, int *year, int *month, int *day)
{
    int y = 4716;
    int j = 1401;
    int m = 2;
    int n = 12;
    int r = 4;
    int p = 1461;
    int v = 3;
    int u = 5;
    int s = 153;
    int w = 2;
    float f = jd+j;
    long e = r*f+v;
    int g = e%p;
    g /= r;
    int h = u*g+w;
    *day = h%s;
    *day /= u;
    *month = (h/s+m)%n;
    *month += 1;
    *year = e/p-y+(n+m-*month)/n;
}

int main() {
    int y = 2011;
    int m = 8;
    int d1 = 23;
    int d2 = 25;
    float j1 = getJulianDay(y,m,d1);
    float j2 = getJulianDay(y,m,d2);
    cout << j2-j1 << endl;
    float j3 = j2+5;
    GetGregorianDay(j3,&y,&m,&d1);
    cout << y << "/" << m << "/" << d1 << endl;
    return 0;
}