无法将字符串文字分配给装箱的std :: string矢量 [英] Can't assign string literal to boxed std::string vector
问题描述
这是我的类型系统的简化版本:
This is a simplified version of my type system:
#include <string>
#include <vector>
template<typename T>
class Box {
public:
Box(const T& value) : _value(value) {};
private:
T _value;
/* ... */
};
typedef Box<int> Int;
typedef Box<double> Double;
typedef Box<std::string> String;
int main(int argc, char* argv[]) {
String a("abc");
std::vector<String> b = { std::string("abc"), std::string("def") };
// error C2664: 'Box<std::string>::Box(const Box<std::string> &)' : cannot convert argument 1 from 'const char' to 'const std::string &'
std::vector<String> c = { "abc", "def" };
}
而 a
和 b
编译, c
不能编译,原因似乎是我尝试从初始化const char
。这就提出了两个问题:
While a
and b
compile, c
does not and the reason seems to be that I try to initialize from const char
. This raises two questions:
-
为什么
b
可能,但c
?是因为std :: vector< Box< std :: string>中的嵌套模板吗? >
?
我可以在不破坏常规的情况下使 c
工作吗拳击机制(参见 typedef Box< double> Double
?
Can I make c
work without destroying the general boxing mechanism (cf. typedef Box<double> Double
?
推荐答案
c
当前需要2次隐式用户转换( const char [N]
-> std :: string
-> String
),而只允许一个。
c
currently need 2 implicit user conversions (const char [N]
-> std::string
-> String
) whereas only one is allowed.
您可以将模板构造函数添加到 Box
You may add template constructor to Box
template<typename T>
class Box {
public:
Box() = default;
Box(const Box&) = default;
Box(Box&&) default;
~Box() = default;
Box& operator=(const Box&) = default;
Box& operator=(Box&&) = default;
template <typename U0, typename ...Us,
std::enable_if_t<std::is_constructible<T, U0, Us...>::value
&& (!std::is_same<Box, std::decay_t<U0>>::value
|| sizeof...(Us) != 0)>* = nullptr>
Box(U0&& u0, Us&&... us) : _value(std::forward<U0>(u0), std::forward<Us>(us)...) {}
private:
T _value;
/* ... */
};
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