当我们声明一个额外的模板参数而不在定义中使用时，为什么编译器会抛出错误？

Question

I have a below code.

#include <iostream>

template <class T,class U>
T myMax(T x, T y)
{
   return (x > y)? x: y;
}

int main()
{
  std::cout << myMax(3, 7) << std::endl;  // Call myMax for int
  std::cout << myMax(3.0, 7.0) << std::endl; // call myMax for double
  std::cout << myMax('g', 'e') << std::endl;   // call myMax for char
  return 0;
}

On compiling the code, compiler reports an error as show below.

functionTemplates.cpp: In function ‘int main()’:
functionTemplates.cpp:18: error: no matching function for call to ‘myMax(int, int)’
functionTemplates.cpp:19: error: no matching function for call to ‘myMax(double, double)’
functionTemplates.cpp:20: error: no matching function for call to ‘myMax(char, char)’

I know that if i remove class U, compilation will be successful. But i want to know why does compiler bothers about an unused parameter?

Answer 1

The compiler cannot determine the type of the unused template argument. You need to specify it explicitly, or remove the unused template argument.

Answer 2

For a general case, a compiler can determine the template parameters from:

1. The arguments used to make the function call.
2. The explicitly used types to make the function call.

In your case, U cannot be determined from the arguments used to make the function call since U is not used by the arguments. The only other way the compiler can determine U is if it is used explicitly in the function call. Eg

 std::cout << myMax<int, double>(3, 7) << std::endl;

---

PS It's not clear to me why you have U as a template parameter in the first place. It's not used at all. Won't it be easier to use:

template <class T>
T myMax(T x, T y)
{
   return (x > y)? x: y;
}