尝试支持文字时构造函数重载歧义
Constructor Overloading Ambiguity When Trying to Support Literals
代码
#include <iostream>
template <typename Type>
class MyContainer
{
private:
Type contained;
public:
MyContainer<Type>(Type & a): contained(a) { std::cout << "&\n"; }
MyContainer<Type>(Type a): contained(a) { std::cout << "_\n"; }
};
class Epidemic
{
private:
int criticality;
public:
Epidemic(int c): criticality(c);
};
int main()
{
// using objects //
Epidemic ignorance(10);
MyContainer<Epidemic> testtube(ignorance); // should print "&"; error instead
// using primitive //
double irrationalnumber = 3.1415;
MyContainer<double> blasphemousnumber(irrationalnumber); // should print "&"; error instead
// using literal //
MyContainer<double> digits(123456789.0); // prints "_"
}
描述
MyContainer<Type>(Type & a)
适用于大多数情况。但是,它不适用于文字(例如 1.732
)。这就是我添加 MyContainer<Type>(Type a)
的原因。但是,通过添加它,我最终会产生歧义,因为非文字可以使用任一构造函数。
问题
Is there a way to satisfy all parameters (both literal and non-literal) given to a constructor?
只需将值类型参数更改为 const
引用:
template <typename Type>
class MyContainer
{
private:
Type contained;
public:
MyContainer<Type>(Type & a): contained(a) { std::cout << "&\n"; }
MyContainer<Type>(const Type& a): contained(a) { std::cout << "_\n"; }
// ^^^^^ ^
};
代码
#include <iostream>
template <typename Type>
class MyContainer
{
private:
Type contained;
public:
MyContainer<Type>(Type & a): contained(a) { std::cout << "&\n"; }
MyContainer<Type>(Type a): contained(a) { std::cout << "_\n"; }
};
class Epidemic
{
private:
int criticality;
public:
Epidemic(int c): criticality(c);
};
int main()
{
// using objects //
Epidemic ignorance(10);
MyContainer<Epidemic> testtube(ignorance); // should print "&"; error instead
// using primitive //
double irrationalnumber = 3.1415;
MyContainer<double> blasphemousnumber(irrationalnumber); // should print "&"; error instead
// using literal //
MyContainer<double> digits(123456789.0); // prints "_"
}
描述
MyContainer<Type>(Type & a)
适用于大多数情况。但是,它不适用于文字(例如 1.732
)。这就是我添加 MyContainer<Type>(Type a)
的原因。但是,通过添加它,我最终会产生歧义,因为非文字可以使用任一构造函数。
问题
Is there a way to satisfy all parameters (both literal and non-literal) given to a constructor?
只需将值类型参数更改为 const
引用:
template <typename Type>
class MyContainer
{
private:
Type contained;
public:
MyContainer<Type>(Type & a): contained(a) { std::cout << "&\n"; }
MyContainer<Type>(const Type& a): contained(a) { std::cout << "_\n"; }
// ^^^^^ ^
};