下面的代码片段是一个游戏,编译器抱怨返回值,所以我想要一些关于如何以其他方式做这个技巧的反馈,让一个函数根据放入的类型返回两种不同的类型但没有超载返回多种类型
template <typename T>
T GetTimeDead(uint64 Guid)
{
bool stringOutput;
if(typeid(T) == typeid(float))
stringOutput = false;
else
stringOutput = true;
bool found = false;
for(map<uint32, TrackInfo>::iterator itr = dieTracker.begin(); itr != dieTracker.end(); ++itr)
{
if(itr->second.GUID == Guid)
{
found = true;
break;
}
}
if(!found)
stringOutput ? return "never" : return sObjectMgr->FindCreature(Guid)->GetCreatureData()->spawntimesecs;
if(!stringOutput)
return dieTracker.find(Guid)->second.seconds;
float seconds = dieTracker.find(Guid)->second.seconds;
uint64 secs = seconds % 60;
uint64 minutes = seconds % 3600/60;
uint64 hours = seconds % 86400/3600;
uint64 days = seconds/86400;
ostringstream ss;
if(days)
days != 1 ? ss << days << " Days " : ss << days << " Day ";
if(hours)
hours != 1 ? ss << hours << " Hours" : ss << hours << " Hour";
if(minutes)
minutes != 1 ? ss << minutes << " Minutes " : ss << minutes << " Minutes ";
if(secs || (!days && !hours && !minutes))
secs != 1 ? ss << secs << " Seconds " : ss << secs << " Second ";
ss << "ago";
return ss.str();
}
什么是错误? – 2013-05-11 19:35:21
错误C2059:语法错误:'return' – user2373581 2013-05-11 19:37:12
它听起来像编译器是正确的:你正在尝试做一件坏事。你可以添加一个解释你准备完成什么? – Elazar 2013-05-11 19:37:37