Question

A puzzling question to C++ template and type conversion is shown below…
To make my life easier, I defined a template class to model a one-to-one relationship using a class BiMap:

#include <stdio.h>
#include <stdlib.h>
#include <map>   
#include <string>

template<class T1, class T2>
class BiMap {
 public:
  T2& operator[] (T1& t1) {
    return d1[t1];
  }
  T1& operator[] (T2& t2) {
    return d2[t2];
  }
 private:
  std::map<T1, T2> d1;
  std::map<T2, T1> d2;
};


int main(int argc, char *argv[])
{

  BiMap<std::string, int> m;
  m["1"] = 2;
  m[2] = 3;

  printf("%d", m["1"]);
  printf("%d", m[2]); 
  return 0;     
}

But I get this compiling error:

testPedigree.cpp:45: error: no match for ‘operator[]’ in ‘m["1"]’
testPedigree.cpp:16: note: candidates are: T2& BiMap<T1, T2>::operator[](T1&) [with T1 = std::basic_string<char, std::char_traits<char>, std::allocator<char> >, T2 = int]
testPedigree.cpp:19: note:                 T1& BiMap<T1, T2>::operator[](T2&) [with T1 = std::basic_string<char, std::char_traits<char>, std::allocator<char> >, T2 = int]

I was expecting C++ will automatically cast const char* to std::string, as I asked the question in this post: Why I can use const char* as key in std::map<std::string, int>

Answer 1

Your code tries to bind a non-const reference to a temporary (namely, the std::string which is implicitly created from your const char[]). Only const references can be bound to temporaries.

Try:

  T2& operator[] (const T1& t1) {
    return d1[t1];
  }
  T1& operator[] (const T2& t2) {
    return d2[t2];
  }

You have some other trivial bugs in your sample. Here is a corrected, tested program:

#include <stdio.h>
#include <stdlib.h>
#include <map>
#include <string>

template<class T1, class T2>
class BiMap {
 public:
  T2& operator[] (const T1& t1) {
    return d1[t1];
  }
  T1& operator[] (const T2& t2) {
    return d2[t2];
  }
 private:
  std::map<T1, T2> d1;
  std::map<T2, T1> d2;
};


int main(int argc, char *argv[])
{

  BiMap<std::string, int> m;
  m["1"] = 2;
  m[2] = "3";

  printf("%d", m["1"]);
  printf("%s\n", m[2].c_str());
  return 0;
}