具有超过1个typename的模板函数

Question

It seems C++ template can automatically deduct the type, when there is only one template parameter. For example,

template<typename T> 

void f(const T &t)
{
  cout << t << endl;
}

f(1);

is fine.

However,

template<typename T1, typename T2> 

void f(const T1 &t1, const T2 &t2)
{
  cout << t1 << t2 << endl;
}

f(1, 2);

the above code can not pass compilation. Is this because of the ability of the compiler or it is against the standard?

Answer 1

The second function and call is perfectly valid. Your compiler is bugged or flat out old rejecting it.