问题
Why does this not work in C++?
Why can't I restrict foo's parameter to std::vector<T>::iterator like this, and what is the best workaround?
#include <vector>
template<class T>
void foo(typename std::vector<T>::iterator) { }
int main()
{
std::vector<int> v;
foo(v.end());
}
The error is:
In function ‘int main()’:
error: no matching function for call to ‘foo(std::vector<int>::iterator)’
note: candidate is:
note: template<class T> void foo(typename std::vector<T>::iterator)
note: template argument deduction/substitution failed:
note: couldn’t deduce template parameter ‘T’
回答1:
The main reason it doesn't work is because the sandard says that
the T here is in a non-deduced context. The reason why the
context is not deduced is because when you pass some type to the
function, the compiler would have to instantiate every single
possible std::vector (including for types not present in this
particular translation unit) in order to try to find one which
had a corresponding type.
Of course, in case of std::vector, the compiler could contain
some magic to make this work, since the semantics of the class
are defined by the standard. But generally,
TemplateClass<T>::NestedType can be a typedef to literally
anything, and there's nothing the compiler can do about it.
回答2:
Simple.
struct X {};
template<> class std::vector<X> {
typedef std::vector<int>::iterator iterator;
};
Oops.
Templates are Turing-Complete. You are asking the compiler to infer the arguments from the results. This is impossible in the general case, even ignoring the possibility of non-one-to-one correspondence.
Typically, you take the iterator type itself as template parameter. This permits other random-access iterators like deque, circular buffer, etc.
来源：https://stackoverflow.com/questions/15208628/why-does-iterator-type-deduction-fail