构造函数的 C++ 模板特化
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C++ template specialization of constructor
提问by user231536
I have a templated class A<T, int> and two typedefs A<string, 20> and A<string, 30>. How do I override the constructor for A<string, 20> ? The following does not work:
我有一个模板类 A<T, int> 和两个 typedef A<string, 20> 和 A<string, 30>。如何覆盖 A<string, 20> 的构造函数？以下不起作用：
template <typename T, int M> class A;
typedef A<std::string, 20> one_type;
typedef A<std::string, 30> second_type;
template <typename T, int M>
class A {
public:
A(int m) {test= (m>M);}
bool test;
};
template<>
one_type::one_type() { cerr << "One type" << endl;}
I would like the class A<std::string,20> to do something that the other class doesn't. How can I do this without changing the constructor A:A(int) ?
我希望 A<std::string,20> 类做一些其他类没有做的事情。如何在不更改构造函数 A:A(int) 的情况下执行此操作？
回答by xtofl
The only thing you cannot do is use the typedefto define the constructor. Other than that, you ought to specialize the A<string,20>constructor like this:
您唯一不能做的就是使用typedef来定义构造函数。除此之外，您应该A<string,20>像这样专门化构造函数：
template<> A<string,20>::A(int){}
If you want A<string,20>to have a differentconstructor than the generic A, you need to specialize the whole A<string,20>class:
如果你想A<string,20>有一个不同于generic 的构造函数A，你需要专门化整个A<string,20>类：
template<> class A<string,20> {
public:
A(const string& takethistwentytimes) { cerr << "One Type" << std::endl; }
};
回答by seh
Assuming your reallymeant for A::testto be publicly accessible, you could do something like this:
假设您真的打算A::test公开访问，您可以执行以下操作：
#include <iostream>
template <int M>
struct ABase
{
ABase(int n) : test_( n > M )
{}
bool const test_;
};
template <typename T, int M>
struct A : ABase<M>
{
A(int n) : ABase<M>(n)
{}
};
template <typename T>
A<T, 20>::A(int n)
: ABase<20>(n)
{ std::cerr << "One type" << std::endl; }
Kick the tires:
踢轮胎：
int main(int argc, char* argv[])
{
A<int, 20> a(19);
std::cout << "a:" << a.test_ << std::endl;
A<int, 30> b(31);
std::cout << "b:" << b.test_ << std::endl;
return 0;
}
回答by amc176
This may be a little bit late, but if you have access to c++11you can use SFINAEto accomplish just what you want:
这可能有点晚了，但是如果您可以访问，则c++11可以使用SFINAE来完成您想要的操作：
template <class = typename std::enable_if<
std::is_same<A<T,M>, A<std::string, 20>>::value>::type // Can be called only on A<std::string, 20>
>
A() {
// Default constructor
}
回答by Franci Penov
You can't with your current approach. one_type is an alias to a particular template specialization, so it gets whatever code the template has.
你不能用你目前的方法。one_type 是特定模板特化的别名，因此它获取模板具有的任何代码。
If you want to add code specific to one_type, you have to declare it as a subclass of A specialization, like this:
如果要添加特定于 one_type 的代码，则必须将其声明为 A 特化的子类，如下所示：
class one_type:
public A<std::string, 20>
{
one_type(int m)
: A<str::string, 20>(m)
{
cerr << "One type" << endl;
}
};
回答by OneOfOne
How about :
怎么样 ：
template<typename T, int M, bool dummy = (M > 20) >
class A {
public:
A(int m){
// this is true
}
};
template<typename T, int M>
class A<T,M,false> {
public:
A(int m) {
//something else
}
};
回答by WaltK
The best solution I've been able to come up with for this situation is to use a "constructor helper function":
对于这种情况，我能够提出的最佳解决方案是使用“构造函数辅助函数”：
template <typename T, int M> class A;
typedef A<std::string, 20> one_type;
typedef A<std::string, 30> second_type;
template <typename T, int M>
class A {
private:
void cons_helper(int m) {test= (m>M);}
public:
A(int m) { cons_helper(m); }
bool test;
};
template <>
void one_type::cons_helper(int) { cerr << "One type" << endl;}
