C ++如何在类型扣除不可能时调用模板化构造函数 [英] C++ how to call templated constructor when type deduction impossible
问题描述
编译此代码没有问题:
struct A
{
template< typename T> ;
void f(int){}
};
A a;
a.f< double>(42);但是,类似的带有模板构造函数的代码不能编译:
$ b
struct A
{
template< typename T>
A(int){}
};
A a< double>(42);
Gcc在最后一行中出现以下错误: 'token
有没有办法使构造函数示例工作?
解决方案没有办法为构造函数显式指定模板,因为你不能为构造函数命名。
根据你想做什么,这可以使用:
#include< iostream>
#include< typeinfo>
struct A
{
template< typename T>
struct with_type {};
template< typename T>
A(int x,with_type< T>)
{
std :: cout< x:<< x<< '\\\
'
<< T:<< typeid(T).name()<< std :: endl;
}
};
int main()
{
A a(42,A :: with_type< double>());
}
利用类型扣除功能欺骗。
这是一个非常不正式的方法,所以可能有更好的方法来做你需要的。
There are no problems compiling this code:
struct A
{
template<typename T>
void f(int) {}
};
A a;
a.f<double>(42);
However, similar code with templated constructor does not compile:
struct A
{
template<typename T>
A(int) {}
};
A a<double>(42);
Gcc gives the following error in the last line: error: unexpected initializer before '<' token
Is there a way to make constructor example work?
解决方案 There is no way to explicitly specify templates for a constructor, as you cannot name a constructor.
Depending on what you're trying to do, this can be used:
#include <iostream>
#include <typeinfo>
struct A
{
template <typename T>
struct with_type {};
template<typename T>
A(int x, with_type<T>)
{
std::cout << "x: " << x << '\n'
<< "T: " << typeid(T).name() << std::endl;
}
};
int main()
{
A a(42, A::with_type<double>());
}
This "cheats" by taking advantage of type deduction.
This is quite unorthodox, though, so there's probably a better way of doing what you need.
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