我有此静态方法:
CString COutlookCalendarSettingsDlg::GetExitCodeAsString(DWORD dwExitCode)
{
using OutlookExitCodesMap = std::map<DWORD, CString>;
OutlookExitCodesMap mapExitCodes;
mapExitCodes.insert(std::pair<DWORD, CString>(1, _T("NoError")));
mapExitCodes.insert(std::pair<DWORD, CString>(-1, _T("CommandLineArguments")));
mapExitCodes.insert(std::pair<DWORD, CString>(-2, _T("BuildingCalendarList")));
mapExitCodes.insert(std::pair<DWORD, CString>(-3, _T("CalendarEventsPathNullEmpty")));
mapExitCodes.insert(std::pair<DWORD, CString>(-4, _T("CalendarEventsPathNotFound")));
mapExitCodes.insert(std::pair<DWORD, CString>(-5, _T("ModeSwitchNotSpecified")));
mapExitCodes.insert(std::pair<DWORD, CString>(-6, _T("ModeSwitchInvalid")));
mapExitCodes.insert(std::pair<DWORD, CString>(-7, _T("AddEventsMWB")));
mapExitCodes.insert(std::pair<DWORD, CString>(-8, _T("AddEventsSRR")));
mapExitCodes.insert(std::pair<DWORD, CString>(-9, _T("SignOut")));
mapExitCodes.insert(std::pair<DWORD, CString>(-10, _T("ReadMWBData")));
mapExitCodes.insert(std::pair<DWORD, CString>(-11, _T("ReadSRRData")));
return mapExitCodes[dwExitCode];
}
现在,我知道我可以将其转变为类中的全局变量,然后将GetExitCodeAsString
方法从此全局变量中返回。
但是如何将地图定义在方法中,但是,只构建一次呢?我不需要继续重建它。因此,它首次被称为它会构造它,随后的时间它将仅返回值。
可以做到吗?
答案很简单,但丑陋:
CString COutlookCalendarSettingsDlg::GetExitCodeAsString(DWORD dwExitCode)
{
using OutlookExitCodesMap = std::map<DWORD, CString>;
static OutlookExitCodesMap mapExitCodes;
if (mapExitCodes.size()==0)
{
mapExitCodes.insert(std::pair<DWORD, CString>(1, _T("NoError")));
mapExitCodes.insert(std::pair<DWORD, CString>(-1, _T("CommandLineArguments")));
mapExitCodes.insert(std::pair<DWORD, CString>(-2, _T("BuildingCalendarList")));
mapExitCodes.insert(std::pair<DWORD, CString>(-3, _T("CalendarEventsPathNullEmpty")));
mapExitCodes.insert(std::pair<DWORD, CString>(-4, _T("CalendarEventsPathNotFound")));
mapExitCodes.insert(std::pair<DWORD, CString>(-5, _T("ModeSwitchNotSpecified")));
mapExitCodes.insert(std::pair<DWORD, CString>(-6, _T("ModeSwitchInvalid")));
mapExitCodes.insert(std::pair<DWORD, CString>(-7, _T("AddEventsMWB")));
mapExitCodes.insert(std::pair<DWORD, CString>(-8, _T("AddEventsSRR")));
mapExitCodes.insert(std::pair<DWORD, CString>(-9, _T("SignOut")));
mapExitCodes.insert(std::pair<DWORD, CString>(-10, _T("ReadMWBData")));
mapExitCodes.insert(std::pair<DWORD, CString>(-11, _T("ReadSRRData")));
}
return mapExitCodes[dwExitCode];
}
此代码对于多线程不安全。另外,如果有未知的出口码,地图就会成长。无需这个...
但是为什么要使用地图作为如此简单的代码。更简单,不使用任何堆,甚至任何结构:
CString COutlookCalendarSettingsDlg::GetExitCodeAsString(DWORD dwExitCode)
{
using OutlookExitCodesMap = std::map<DWORD, CString>;
static const struct {
int dwCode;
LPCTSTR pszText;
}
aMap[] =
{
1, _T("NoError"),
-1, _T("CommandLineArguments"),
-2, _T("BuildingCalendarList"),
-3, _T("CalendarEventsPathNullEmpty"),
-4, _T("CalendarEventsPathNotFound"),
-5, _T("ModeSwitchNotSpecified"),
-6, _T("ModeSwitchInvalid"),
-7, _T("AddEventsMWB"),
-8, _T("AddEventsSRR"),
-9, _T("SignOut"),
-10, _T("ReadMWBData"),
-11, _T("ReadSRRData"),
};
for (const auto &data : aMap)
{
if (static_cast<DWORD>(data.dwCode)==dwExitCode)
return data.pszText;
}
return CString();
}
代码可能具有错别字...我刚刚从头开始写
这是基于我问题提供的评论的另一个解决方案:
CString COutlookCalendarSettingsDlg::GetExitCodeAsString(DWORD dwExitCode)
{
switch (dwExitCode)
{
case 1: return _T("NoError");
case -1: return _T("CommandLineArguments");
case -2: return _T("BuildingCalendarList");
case -3: return _T("CalendarEventsPathNullEmpty");
case -4: return _T("CalendarEventsPathNotFound");
case -5: return _T("ModeSwitchNotSpecified");
case -6: return _T("ModeSwitchInvalid");
case -7: return _T("AddEventsMWB");
case -8: return _T("AddEventsSRR");
case -9: return _T("SignOut");
case -10: return _T("ReadMWBData");
case -11: return _T("ReadSRRData");
}
return CString();
}