在模板,在那里,为什么我必须把typename和template上依赖的名字呢?究竟什么是依赖名称?我有以下代码:
template <typename T, typename Tail> // Tail will be a UnionNode too.
struct UnionNode : public Tail {
// ...
template<typename U> struct inUnion {
// Q: where to add typename/template here?
typedef Tail::inUnion<U> dummy;
};
template< > struct inUnion<T> {
};
};
template <typename T> // For the last node Tn.
struct UnionNode<T, void> {
// ...
template<typename U> struct inUnion {
char fail[ -2 + (sizeof(U)%2) ]; // Cannot be instantiated for any …Run Code Online (Sandbox Code Playgroud) 可能重复:
我必须在何处以及为何要使用"template"和"typename"关键字?
c ++模板typename迭代器
以下代码将无法编译:
#include <iostream>
#include <set>
using namespace std;
template<class T>
void printSet(set<T> s){
set<T>::iterator it;
}
int main(int argc, char** argv){
set<int> s;
printSet<int>(s);
return 0;
}
Run Code Online (Sandbox Code Playgroud)
我收到一个错误说:
set.cpp: In function ‘void printSet(std::set<T, std::less<_Key>, std::allocator<_CharT> >)’:
set.cpp:7: error: expected `;' before ‘it’
set.cpp: In function ‘void printSet(std::set<T, std::less<_Key>, std::allocator<_CharT> >) [with T = int]’:
set.cpp:12: instantiated from here
set.cpp:7: error: dependent-name ‘std::set<T,std::less<_Key>,std::allocator<_CharT> >::iterator’ is parsed as a non-type, but instantiation yields a type
set.cpp:7: note: say …Run Code Online (Sandbox Code Playgroud)