我正在尝试让一个函数采用通用 std::vector (template<typename T> std::vector<T>
,然后调用一个模板函数,该函数在其所有元素上都具有针对特定(抽象)类型的特化。
我正在尝试弄清楚如何使用专用版本,同时仍然能够使用通用版本,但我没有成功。
我使用的是 visual studio 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:\...\MSVC\14.20.27508\include\ostream(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:\...\MSVC\14.20.27508\include\ostream(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 ==========```
最佳答案
为了选择专业 const Foo&
, 从 foo_impl
隐式转换至 Foo
应该发生。另一方面,通过选择 const T&
编译器做了一个聪明的举动,它直接推导出类型(即 foo_impl
),因为这对编译器来说更容易。
因此 Foo
的特化被拒绝,错误出现是因为未找到 operator<<(std::cout, foo_impl)
的过载.
更新:为了选择正确的特化,通过阻止编译器实例化 const T&
对于 foo_impl
,您可以将 SFINAE 与重载解析一起使用。 (创作者@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;
}
关于c++ - 是否可以重载模板函数以与 std::vector 的元素一起使用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55771614/