从函数返回 ::iterator 不编译

Question

以下 repo 编译正常：

#include <tuple>
#include <vector>

class Foo
{
public:

    typedef std::tuple<int, int> foo_type;

    std::vector<foo_type>::iterator Find() {
        return Listeners.begin();
    }

    std::vector<foo_type> Listeners;
};

int main() 
{
    Foo foo;

    auto i = foo.Find();
}

但是当我在 P、Q 上模板化元组时，如以下 repo 所示，我收到此错误：

syntax error : missing ';' before identifier 'Find'

#include <tuple>
#include <vector>

template<typename P, typename Q>
class Foo
{
public:

    typedef std::tuple<P, Q> foo_type;

    std::vector<foo_type>::iterator Find() {
        return Listeners.begin();
    }

    std::vector<foo_type> Listeners;
};

int main()
{
    Foo<int, int> foo;

    auto i = foo.Find();
}

Answer 1

您在这里缺少 typename：

typename std::vector<foo_type>::iterator Find() {
^^^^^^^^
    return Listeners.begin();
}

有关更详细的说明，请参阅 Where and why do I have to put the "template" and "typename" keywords?

从函数返回 ::iterator 不编译

Returning ::iterator from function does not compile

c++

templates

tuples