Confusing Template error

问题:

I've been playing with clang a while, and I stumbled upon "test/SemaTemplate/dependent-template-recover.cpp" (in the clang distribution) which is supposed to provide hints to recover from a template error.

The whole thing can be easily stripped down to a minimal example:

template<typename T, typename U, int N> struct X {
    void f(T* t)
    {
        // expected-error{{use 'template' keyword to treat 'f0' as a dependent template name}}
        t->f0<U>();
    }
};

The error message yielded by clang:

tpl.cpp:6:13: error: use 'template' keyword to treat 'f0' as a dependent template name
         t->f0<U>();
            ^
            template 
1 error generated.

... But I have a hard time understanding where exactly one is supposed to insert the template keyword to have the code to be syntactically correct?

回答1:

ISO C++03 14.2/4:

When the name of a member template specialization appears after . or -> in a postfix-expression, or after nested-name-specifier in a qualified-id, and the postfix-expression or qualified-id explicitly depends on a template-parameter (14.6.2), the member template name must be prefixed by the keyword template. Otherwise the name is assumed to name a non-template.

In t->f0<U>(); f0<U> is a member template specialization which appears after -> and which explicitly depends on template parameter U, so the member template specialization must be prefixed by template keyword.

So change t->f0<U>() to t->template f0<U>().

回答2:

In addition to the points others made, notice that sometimes the compiler couldn't make up his mind and both interpretations can yield alternative valid programs when instantiating

#include <iostream>

template<typename T>
struct A {
  typedef int R();

  template<typename U>
  static U *f(int) { 
    return 0; 
  }

  static int f() { 
    return 0;
  }
};

template<typename T>
bool g() {
  A<T> a;
  return !(typename A<T>::R*)a.f<int()>(0);
}


int main() {
  std::cout << g<void>() << std::endl;
}

This prints 0 when omitting template before f<int()> but 1 when inserting it. I leave it as an exercise to figure out what the code does.

回答3:

Insert it just before the point where the caret is:

template<typename T, typename U, int N> struct X {
     void f(T* t)
     {
        t->template f0<U>();
     }
};

Edit: the reason for this rule becomes clearer if you think like a compiler. Compilers generally only look ahead one or two tokens at once, and don't generally "look ahead" to the rest of the expression.[Edit: see comment] The reason for the keyword is the same as why you need the typename keyword to indicate dependent type names: it's telling the compiler "hey, the identifier you're about to see is the name of a template, rather than the name of a static data member followed by a less-than sign".

回答4:

Excerpt from C++ Templates

The .template Construct A very similar problem was discovered after the introduction of typename. Consider the following example using the standard bitset type:

template<int N> 
void printBitset (std::bitset<N> const& bs) 
{ 
    std::cout << bs.template to_string<char,char_traits<char>, 
                                       allocator<char> >(); 
} 

The strange construct in this example is .template. Without that extra use of template, the compiler does not know that the less-than token (<) that follows is not really "less than" but the beginning of a template argument list. Note that this is a problem only if the construct before the period depends on a template parameter. In our example, the parameter bs depends on the template parameter N.

In conclusion, the .template notation (and similar notations such as ->template) should be used only inside templates and only if they follow something that depends on a template parameter.