error: no matching function for call to sort

Question

From Matthew H. Austern's STL book, "Block" class for a fixed size array.

template<class T, size_t N>
struct Block
{
   // Various typedef's

   // Implementations of begin() and end().

   reference operator[](size_t nIndex)
   {
     return data[nIndex];
   }

   // I added this overloaded operator+
   T* operator+(const size_t nIncrement)
   {
     return (data + nIncrement);
   }

   // I added this overloaded cast operator
   // **Why is this cast operator not being called?**
   operator T*()
   {
     return data;
   }

   T data[N];
};

This is what I am trying to do:

Block<int, 10> tArray; // Populate tArray.

// I want to use std::sort() in the same way as I would
// for a regular array.
sort(tArray, tArray + tArray.size());

I get the compiler error:

error: no matching function for call to sort(Block&, int*)

Why isn't the compiler aware of the overloaded cast operator?

All of the following compile:

sort(tArray + 0, tArray + tArray.size());
sort(static_cast<int*>(tArray), tArray + tArray.size());
sort(tArray.begin(), tArray.end());

Obviously, there is something about how overloaded cast operators work that I am ignorant of. Any ideas?

Thanks.

CONSTRAINTS:

I may not use a C++11 (so no std::array from C++11). I may not use begin() nor end().

Answer 1

Your requirement that you may use begin() nor end() is problematic. Sort requires random access iterator; see Bjarne Stroustrup The C++ programming language 4^th edition and std::sort.

Using an explicit cast and pointer artithmic, you should be able to do the following:

For example let T be std::string, let N be 42, and let b an initialised Block<T,N>:

Block<std::string, 42> b;
std::string* begin = (std::string*)b;
std::string* end = begin + 42; // Or if Block has Block.size()
                               // std::string* end = begin + b.size()

std::sort(begin, end);

Now b contains the strings sorted according to bool operator<(const std::string&, const std::string&).