上一个问题:std::string 类继承和繁琐的 c++ 重载解析
在上一个问题之后的步骤中,我尝试测试operator+
原始字符串指针:"aaa" + path_string{ "bbb" }
. 并发现它没有调用相应的path_string
类友函数。
我试图添加不模板重载operator+
(2)
,但它也没有工作。但我发现模板化的(3)
确实有效。
#include <string>
template <class t_elem, class t_traits, class t_alloc>
class path_basic_string : public std::basic_string<t_elem, t_traits, t_alloc>
{
public:
using base_type = std::basic_string<t_elem, t_traits, t_alloc>;
path_basic_string() = default;
path_basic_string(const path_basic_string & ) = default;
path_basic_string(path_basic_string &&) = default;
path_basic_string & operator =(path_basic_string path_str)
{
this->base_type::operator=(std::move(path_str));
return *this;
}
path_basic_string(base_type r) :
base_type(std::move(r))
{
}
path_basic_string(const t_elem * p) :
base_type(p)
{
}
base_type & str()
{
return *this;
}
const base_type & str() const
{
return *this;
}
using base_type::base_type;
using base_type::operator=;
// ... all over operators are removed as not related to the issue ...
// (1)
friend path_basic_string operator+ (const t_elem * p, const base_type & r)
{
path_basic_string l_path = p;
l_path += "xxx";
return std::move(l_path);
}
friend path_basic_string operator+ (const t_elem * p, base_type && r)
{
if (!r.empty()) {
return "111" + ("/" + r); // call base operator instead in case if it is specialized for this
}
return "111";
}
// (2)
friend path_basic_string operator+ (const t_elem * p, path_basic_string && r)
{
base_type && r_path = std::move(std::forward<base_type>(r));
if (!r_path.empty()) {
return "222" + ("/" + r_path); // call base operator instead in case if it is specialized for this
}
return "222";
}
// (3) required here to intercept the second argument
template <typename T>
friend path_basic_string operator+ (const t_elem * p, T && r)
{
base_type && r_path = std::move(std::forward<base_type>(r));
if (!r_path.empty()) {
return "333" + ("/" + r_path); // call base operator instead in case if it is specialized for this
}
return "333";
}
};
using path_string = path_basic_string<char, std::char_traits<char>, std::allocator<char> >;
std::string test_path_string_operator_plus_right_xref(path_string && right_path_str)
{
return "aaa" + right_path_str;
}
int main()
{
const path_string test =
test_path_string_operator_plus_right_xref(std::move(path_string{ "bbb" }));
printf("-%s-\n", test.str().c_str());
return 0;
}
3 个编译器的输出:gcc 5.4、clang 3.8.0、msvc 2015 (19.00.23506)
-333/bbb-
https://rextester.com/BOFUS59590
正如我所记得的,C++ 标准澄清了这一点,因为只有当没有一个非模板化函数与参数完全匹配时,才需要查找模板化函数。但是(2)
操作符必须完全匹配,但是为什么它甚至没有被调用呢?
如果 remove(3)
则将(1)
调用而不是(2)
which is match than (1)
。
这里发生了什么?
PS:我认为这与上一个问题中的const
+single reference
类似。