c ++函数模板指定第二个模板参数类型

Question

just started learning template programming, I have the following code,

template<typename T, typename U>
void add(T x, U y)
{
    cout<< x + y <<endl;
}

I can call this by,

add(1, 2);
add<int, int>(1, 2);
add<int>(1, 2.0);

In the third case, I believe that means I specified [T=int] , and compiler will deduce [U=double]

My question is how can I specify the second parameter type explicitly?

Answer 1

What about add(1,(int)2.0); .

In theory according to template argument deduction rules, this causes the second template parameter to be deduced as int .So this is stricktly equivalent to this hypothetical syntax add<U=int>(1,2.0);

So this is the way to specify the second template argument!

It is impossible to find an equivalent syntax if the second template argument is non-deductible:

template<class T>
struct t_{
  using type = T;
  };

template<class T,class U>
auto add(T, typename t_<U>::type);