为什么隐式类型转换在模板扣除中不起作用？

发布于 2025-02-02 13:54:15 字数 1003 浏览 2 评论 0原文

在以下代码中，我想通过将int将其转换为scalar＆lt; int＆gt;对象来调用模板函数。

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar{
public:
  Scalar(Dtype v) : value_(v){}
private:
  Dtype value_;
};

template<typename Dtype>
void func(int a, Scalar<Dtype> b){ 
  cout << "ok" <<endl;
}

int main(){
  int a = 1;
  func(a, 2); 
  //int b = 2;
  //func(a, b);
  return 0;
}

为什么模板参数扣除/替换失败？评论的编码也是错误的。

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
   func(a, 2);
            ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
      ^
test.cpp:13:6: note:   template argument deduction/substitution failed:
test.cpp:19:12: note:   mismatched types ‘Scalar<Dtype>’ and ‘int’
   func(a, 2);

原文

In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar{
public:
  Scalar(Dtype v) : value_(v){}
private:
  Dtype value_;
};

template<typename Dtype>
void func(int a, Scalar<Dtype> b){ 
  cout << "ok" <<endl;
}

int main(){
  int a = 1;
  func(a, 2); 
  //int b = 2;
  //func(a, b);
  return 0;
}

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
   func(a, 2);
            ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
      ^
test.cpp:13:6: note:   template argument deduction/substitution failed:
test.cpp:19:12: note:   mismatched types ‘Scalar<Dtype>’ and ‘int’
   func(a, 2);

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薯片软お妹 2025-02-09 13:54:15

因为 template参数ducation 不是那么聪明：它不是（按设计），考虑用户定义的转换。和int - ＆gt; 标量＆lt; int＆gt;是用户定义的转换。

如果要使用TAD，则需要在呼叫者站点上转换参数：

func(a, Scalar<int>{2});

或定义扣除指南¹scalar 并调用f ：

func(a, Scalar{2}); // C++17 only

另外，您可以明确实例化f（仅当scalar＆lt; t＆gt; :: starcar（t） is not equarlicit ）：

func<int>(a, 2);

^1）默认值扣除指南就足够了： demo 。

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>{2});

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar{2}); // C++17 only

Alternatively, you can explicitly instantiate f (only works if Scalar<T>::Scalar(T) is not explicit):

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

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紫竹語嫣☆ 2025-02-09 13:54:15

template<class Dtype>
void func(int a, Scalar<Dtype> b){ 
  std::cout << "ok" << std::endl;
}
template<class Dtype>
void func(int a, Dtype b){ 
  func(a, Scalar<Dtype>(std::move(b)));
}

模板参数扣除是模式匹配，它仅与完全匹配的类型或基本类型。它没有转换。

转换以后进行，以分辨率分辨率＆amp;功能致电时间。

在这里，我们将另一个明确将其转发到您想要的过载。

template<class Dtype>
void func(int a, Scalar<Dtype> b){ 
  std::cout << "ok" << std::endl;
}
template<class Dtype>
void func(int a, Dtype b){ 
  func(a, Scalar<Dtype>(std::move(b)));
}

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

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~没有更多了~