递归AsString（）以在C ++中打印STL容器

Question

I'm trying to write an AsString() function that converts STL containers to string according to my taste. Here's the code I've come up with so far:

template<class T>
inline string AsString(const T& v);

template<class First, class Second>
inline string AsString(const pair<First, Second>& p);

template<class Iter>
inline string PrintSequence(const char* delimiters, Iter begin, Iter end) {
  string result;
  result += delimiters[0];
  int size = 0;
  for (size = 0; begin != end; ++size, ++begin) {
    if (size > 0) {
      result += ", ";
    }
    result += AsString(*begin);
  }
  result += delimiters[1];
  result += StringPrintf("<%d>", size);
  return result;
}

#define OUTPUT_TWO_ARG_CONTAINER(Sequence) \
template<class T1, class T2> \
inline string AsString(const Sequence<T1, T2>& seq) { \
  return PrintSequence("[]", seq.begin(), seq.end()); \
}

OUTPUT_TWO_ARG_CONTAINER(vector)
OUTPUT_TWO_ARG_CONTAINER(deque)
OUTPUT_TWO_ARG_CONTAINER(list)

template<class First, class Second>
inline string AsString(const pair<First, Second>& p) {
  return "(" + AsString(p.first) + ", " + AsString(p.second) + ")";
}

template<class T>
inline string AsString(const T& v) {
  ostringstream s;
  s << v;
  return s.str();
}

As you can see, the basic idea is that AsString() recursively calls itself on STL containers and then it bottoms out to the usual operator<<() (the reason I don't want to override operator<<() is because I don't want to interfere with other libraries that do exactly that).

Now, AsString() compiles and works on shallow containers, but not on nested ones:

vector<int> v;
v.push_back(1);
v.push_back(2);
AsString(v) == "[1, 2]<2>";  // true

vector<vector<int> > m;
m.push_back(v);
m.push_back(v);
AsString(m) == "[[1, 2]<2>, [1, 2]<2>]<2>";  // Compilation Error!!!

The compiler, for some reason, wants to use operator<<() when trying to print the elements of `m', despite the fact that I have provided a template specialization for vectors.

How could I make AsString() work?

UPDATE : OK, turns out the order of definitions do matter (at least for this compiler -- gcc 4.4.3). When I put the macro definitions first, the compiler will correctly pick them up and display a vector of vectors. Inexplicable.

Answer 1

The template world is wonderful... and a real trap to the unwary...

A specialization, is taking an existing template function and specifying all its arguments.

An overload is reusing the same name that another function (whether template or not) for a different set of arguments.

template <typename T>
void foo(T const& t);

template <>
void foo<int>(int i); // this is a "complete" specialization

template <typename T, typename U>
void foo<std::pair<T,U>>(std::pair<T,U> const& pair);
  // this is a "partial" specialization
  // and by the way... it does NOT COMPILE

template <typename T, typename U>
void foo(std::pair<T,U> const& pair); // this is an overload

Note the syntactical difference, in the overload there is no <xxxx> after the identifier ( foo here).

It is not possible, in C++, to partially specialize a function; that is to leave some genericity in the arguments. You can either overload or fully specialize: mandatory reading at this point GotW #49: Template Specialization and Overloading

Therefore, the choice is between:

template <typename T>
std::string AsString(const T& v); // (1)

template <typename T, typename Allocator>
std::string AsString(std::vector<T, Allocator> const& v); // (2)

And the real question is: what is the type of *begin ?

Well, m is not const-qualified:

• Iter logically is std::vector< std::vector<int> >::iterator .
• the type of *begin is thus std::vector<int>&

So the two overloads are considered with:

• (1): T = std::vector<int> , requires a conversion to const-ref
• (2): T = int, U = std::allocator<int> , requires a conversion to const-ref

The second should be selected because it's closer to the real type, as far as I understand. I tested it with VC++ 2010 and it actually got selected.

Could you also declare a non-const qualified version of the vector overload and see if it appeases your compiler ? (which I'd like to know the name of, by the way ;) ).

Answer 2

You have not provided a specialization, you have overloaded AsString. As it happens, your later overload isn't preferred over the T const& version.

Instead, overload op<< in a special namespace for the various stdlib containers. The namespace is important so you don't affect other code, but you will explicitly use it in AsString:

namespace make_sure_to_put_these_overloads_in_a_namespace {

// Your PrintSequence adapted to a stream instead of a string:
template<class Iter>
void PrintSequence(std::ostream &s, const char* delim,
                   Iter begin, Iter end)
{
  s << delim[0];
  int size = 0;
  if (begin != end) {
    s << *begin;
    ++size;
    while (++begin != end) {
      s << ", " << *begin;
      ++size;
    }
  }
  s << delim[1] << '<' << size << '>';
}

#define OUTPUT_TWO_ARG_CONTAINER(Sequence) \
template<class T1, class T2> \
std::ostream& operator<<(std::ostream &s, Sequence<T1, T2> const &seq) { \
  PrintSequence(s, "[]", seq.begin(), seq.end()); \
  return s; \
}

OUTPUT_TWO_ARG_CONTAINER(std::vector)
OUTPUT_TWO_ARG_CONTAINER(std::deque)
OUTPUT_TWO_ARG_CONTAINER(std::list)
// other types
#undef OUTPUT_TWO_ARG_CONTAINER

template<class First, class Second>
std::ostream& operator<<(std::ostream &s, std::pair<First, Second> const &p) { \
  s << "(" << p.first << ", " << p.second << ")";
  return s;
}

}

template<class T>
std::string AsString(T const &v) {
  using namespace make_sure_to_put_these_overloads_in_a_namespace;
  std::ostringstream ss;
  ss << v;
  return ss.str();
}