是否可以重载模板函数以与std :: vector的元素一起使用？

Posted 2021-03-26

我试图让一个函数接受一个通用的std::vector (template<typename T> std::vector<T>，然后调用一个模板函数，它对所有元素都有一个特定（抽象）类型的特化。

我试图找出如何使用专用版本，同时仍然能够使用通用版本，但我没有成功。

我正在使用视觉工作室2019。

这是我的代码：

#include <iostream>
#include <vector>

//the abstract type
struct Foo
{
    virtual int bar(unsigned int) const = 0;
};

//The 'template function' I mentioned
template<typename T>
void do_something_with_an_element(const T& value)
{
    std::cout << value;
}

//the specialised version I mentioned
template<>
void do_something_with_an_element<Foo>(const Foo& value)
{
    //Note: this is only a placeholder
    std::cout << "It's a foo. bar = " << value.bar(7) << std::endl;
}

//the function that takes the 'generic' std::vector
template<typename T>
void do_something_with_a_vector(const std::vector<T>& vec)
{
    for (auto element : vec)
    {
        //calling the function on all its elements
        do_something_with_an_element(element); //Here is the problem line
    }
}

struct foo_impl : public Foo
{
    int val;

    foo_impl(int _val)
        : val(_val)
    {}

    // Inherited via Foo
    virtual int bar(unsigned int _something) const override
    {
        //do whatever...
        return val;
    }
};

std::vector<int> int_vector = { 32, 3, 43, 23 }; 
std::vector<foo_impl> foo_vector = { foo_impl(3), foo_impl(9), foo_impl(13) };

int main()
{
    do_something_with_a_vector(int_vector); //fine
    do_something_with_a_vector(foo_vector); //compile error
}

这是我的错误：

        1>program.cpp
        1>C:...program.cpp(17): error C2679: binary '<<': no operator found which takes a right-hand operand of type 'const T' (or there is no acceptable conversion)
        1>        with
        1>        [
        1>            T=foo_impl
        1>        ]
        1>C:...MSVC14.20.27508includeostream(438): note: could be 'std::basic_ostream<char,std::char_traits<char>> &std::basic_ostream<char,std::char_traits<char>>::operator <<(std::basic_streambuf<char,std::char_traits<char>> *)'
        1>C:...MSVC14.20.27508includeostream(413): note: or       'std::basic_ostream<char,std::char_traits<char>> &std::basic_ostream<char,std::char_traits<char>>::operator <<(const void *)'
         ((list goes on...))
        1>C:...program.cpp(17): note: while trying to match the argument list '(std::ostream, const T)'
        1>        with
        1>        [
        1>            T=foo_impl
        1>        ]
        1>C:...program.cpp(35): note: see reference to function template instantiation 'void do_something_with_an_element<foo_impl>(const T &)' being compiled
        1>        with
        1>        [
        1>            T=foo_impl
        1>        ]
        1>C:...program.cpp(61): note: see reference to function template instantiation 'void do_something_with_a_vector<foo_impl>(const std::vector<foo_impl,std::allocator<_Ty>> &)' being compiled
        1>        with
        1>        [
        1>            _Ty=foo_impl
        1>        ]
        1>Done building project "program.vcxproj" -- FAILED.
        ========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========```

答案

为了选择const Foo&的特化，应该发生从foo_impl到Foo的隐式转换。另一方面，通过选择const T&，编译器做了一个巧妙的移动，它直接推导出类型（即foo_impl），因为它对编译器来说更容易。

因此Foo的专业化被拒绝，并且错误出现了operator<<(std::cout, foo_impl)没有发现过载。

---

更新：为了选择正确的专业化，通过阻止编译器为const T&实例化foo_impl，您可以使用SFINAE以及重载决策。 （creadits @Hiroki）

template<typename T>
std::enable_if_t<!std::is_base_of_v<Foo, T>>
do_something_with_an_element(const T& value)
{
    std::cout << value;
}

void do_something_with_an_element(const Foo& value)
{
    std::cout << "It's a foo. bar = " << value.bar(7) << std::endl;
}