add_month從boost :: gregorian中刪除？

Question

我正在使用boost::gregorian來執行日期計算。我想用add_month具體根據例子（電流的1.63 http://www.boost.org/doc/libs/1_63_0/doc/html/date_time/examples.html）

/* Simple program that uses the gregorian calendar to progress by exactly 
* one month, irregardless of how many days are in that month. 
* 
* This method can be used as an alternative to iterators 
*/ 
#include "boost/date_time/gregorian/gregorian.hpp" 
#include <iostream> 

int main() 
{ 

    using namespace boost::gregorian; 

    date d = day_clock::local_day(); 
    add_month mf(1); 
    date d2 = d + mf.get_offset(d); 
    std::cout << "Today is: " << to_simple_string(d) << ".\n" 
    << "One month from today will be: " << to_simple_string(d2) 
    << std::endl; 

    return 0; 
} 

然而，這提供錯誤消息

month.cpp: In function `int main()':         
month.cpp:33:5: error: `add_month' was not declared in this scope  
add_month mf(1);             
^                 
month.cpp:35:19: error: `mf' was not declared in this scope    
date d2 = d + mf.get_offset(d);         
      ^             

Answer 1

事實上。這個例子已經過時了。事實上，我不記得看到這個功能，所以它可能會過時。

我推薦以下方法代替：

/* Simple program that uses the gregorian calendar to progress by exactly 
* one month, irregardless of how many days are in that month. 
* 
* This method can be used as an alternative to iterators 
*/ 

#include "boost/date_time/gregorian/gregorian.hpp" 
#include "boost/date_time/gregorian/gregorian.hpp" 
#include <iostream> 

int main() 
{ 

    using namespace boost::gregorian; 

    date d = day_clock::local_day(), 
     prev = d - months(1), 
     next = d + months(1); 

    std::cout << "Today is: "      << to_simple_string(d) << ".\n" 
       << "One month before today was: " << to_simple_string(prev) << "\n" 
       << "One month from today will be: " << to_simple_string(next) << "\n"; 
} 

打印了（對我來說）：

Today is: 2017-Mar-23. 
One month before today was: 2017-Feb-23 
One month from today will be: 2017-Apr-23