模板函式中使用未知型別的容器--如何遍歷一個未知容器
阿新 • • 發佈：2019-02-14
template<typename T1>
void printList(const T1 & t1)
{
for(typename T1::const_iterator it =t1.begin(); it!=t1.end(); ++it)
{
cout<<(*it) << endl;
}
};如果寫得更好，就得過載 operation << 了。
請注意 typename T1::const_iterator it ;一定要加上typename。即使有些編譯器讓你通過。
不寫的話可能有以下錯誤：
main main.cpp:103: error: ‘it’ was not declared in this scopemain.cpp:91: instantiated from heremain.cpp:103: error: dependent-name ‘std::map<T,A,std::less<_Key>,std::allocator<std::pair<const _Key, _Tp> > >::const_iterator’ is parsed as a non-type, but instantiation yields a type main.cpp:103: note: say ‘typename std::map<T,A,std::less<_Key>,std::allocator<std::pair<const _Key, _Tp> > >::const_iterator’ if a type is mean
如下有問題：
template<class T, class A> void ShowMap(const map<T, A>& v) { for (map<T, A> ::const_iterator ci = v.begin();ci != v.end(); ++ci) cout << ci ->first <<": " << ci ->second <<endl; cout << endl; }
解釋：It accurately doesn't treat map<T, A>::const_iterator as a type because it relies on template parameters, T and A. To make the compiler believe you, you need to use the typename keyword.