模板构造函数的奇怪之处
可能的重复：
可以显式指定构造函数的模板参数吗？< /a>
跟进我之前的问题， （我在编辑2中发现了这种情况）
代码简单：
#include <iostream>
struct Printer
{
Printer() { std::cout << "secret code" << std::endl; }
};
template <class A>
struct Class
{
template <class B, class C>
Class(B arg)
{
C c; /* the 'secret code' should come from here */
std::cout << arg << std::endl;
}
Class(double arg) { std::cout << "double" << std::endl; }
Class(float arg) { std::cout << "float" << std::endl; }
/* this forbids the use of printer in the first parameter */
Class(Printer printer) { throw std::exception(); /* here be dragons */ }
};
int main()
{
Class<int> c(1.0f);
Class<int>* ptr = new Class<int>((double)2.0f);
return 0;
}
// Can anyone print 'secret code' while creating an object of type 'Class' ?
详细：对于模板构造函数，当实例化对象时，您可以指定一个不属于构造函数参数的模板参数吗？
我认为这本身就值得一个问题。
Possible Duplicate:
Can the template parameters of a constructor be explicitly specified?
following up on my previous question, (I found this situation in edit 2)
Laid out simple in code:
#include <iostream>
struct Printer
{
Printer() { std::cout << "secret code" << std::endl; }
};
template <class A>
struct Class
{
template <class B, class C>
Class(B arg)
{
C c; /* the 'secret code' should come from here */
std::cout << arg << std::endl;
}
Class(double arg) { std::cout << "double" << std::endl; }
Class(float arg) { std::cout << "float" << std::endl; }
/* this forbids the use of printer in the first parameter */
Class(Printer printer) { throw std::exception(); /* here be dragons */ }
};
int main()
{
Class<int> c(1.0f);
Class<int>* ptr = new Class<int>((double)2.0f);
return 0;
}
// Can anyone print 'secret code' while creating an object of type 'Class' ?
Detailed: For a template constructor, can you specify a template argument which is not part of the constructor's arguments when an object get's instantiated?
I think this deserves a question of its own.
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评论（2）
不，这是不可能的。
没有任何语法可以为构造函数模板提供显式模板参数。您只能为整个类模板提供显式模板参数。
以下来自
[temp.arg.explicit]的文本（2003 年措辞，14.8.1/5）涵盖了该场景。尽管该子句是非规范性的，但它向我们解释，作为语法的固有限制，这是不可能的：这在一定程度上是由于您自己从未真正显式调用构造函数这一事实。例如，当您编写
A()时，您并没有像函数一样调用构造函数，即使它看起来好像是（“调用转换成员函数模板和构造函数成员函数模板时不使用函数名称”）。No, it's not possible.
There is no syntax with which you can provide explicit template parameters to a constructor template. You can only provide explicit template parameters for the class template as a whole.
The following text from
[temp.arg.explicit](2003 wording, 14.8.1/5) covers the scenario. Though the clause is non-normative, it serves to explain to us that, as an inherent restriction of the grammar, this is not possible:This, partially, comes out of the fact that you never actually invoke the constructor explicitly yourself. When you write, say,
A()you are not calling the constructor like a function, even though it looks as if you are ("conversion member function templates and constructor member function templates are called without using a function name").我想他想知道如何用 C 实例化这个类作为 SomeType：
我不知道这是否可能。
I think he want to know how to instantiate this class with C as SomeType:
I don't know if this is possible.