Question

如何修改此代码以便能够正确解析此字符串＆＃34; 22/12/2016 01:12:10 ＆＃34;到 date_time 变量？

struct tm date_time = {};
std::wstring dateTimeFormat{ L"%d/%m/%Y %H:%M:%S" };
std::istringstream ss("22/12/2016 01:12:10");
ss >> std::get_time(&date_time, dateTimeFormat.c_str());


 1>c:\program files (x86)\microsoft visual studio\2017\community\vc\tools
\msvc\14.12.25827\include\iomanip(190): error C2338: wrong character type for 
get_time
 1>c:\...\procedure.cpp(70): note: see reference to function template 
instantiation 'std::basic_istream<char,std::char_traits<char>> &std::operator 
>><char,std::char_traits<char>,_Elem>(std::basic_istream<char,std::char_traits<char>>
 &,const std::_Timeobj<_Elem,tm    *> &)' being compiled
 1>                   with
 1>                    [
 1>                     _Elem=wchar_t
 1>                     ]
 1>Done building project "projectname" -- FAILED.

Answer 1

您正在将宽字符与窄字符混合。

对于宽字符，代码应为：

    struct tm           date_time = {};
    std::wstring        dateTimeFormat{L"%d/%m/%Y %H:%M:%S"};
    std::wistringstream ss(L"22/12/2016 01:12:10");

    ss >> std::get_time( &date_time, dateTimeFormat.c_str() );

对于狭窄的字符，它应该是：

    struct tm          date_time = {};
    std::string        dateTimeFormat{"%d/%m/%Y %H:%M:%S"};
    std::istringstream ss("22/12/2016 01:12:10");

    ss >> std::get_time( &date_time, dateTimeFormat.c_str() );

无论哪种方式都需要以下标题：

#include <iomanip>
#include <sstream>

std :: get_time中日期时间的格式

1 个答案: