Question

Inspired by this answer, I tried next example :

#include <map>
#include <string>
#include <iostream>

int main()
{
  const std::map< int, std::string > mapping = {
      1, "ONE",
      2, "TWO",
    };

  const auto it = mapping.find( 1 );
  if ( mapping.end() != it )
  {
    std::cout << it->second << std::endl;
  }
  else
  {
    std::cout << "not found!" << std::endl;
  }
}

and the compilation failed with next error message (g++ 4.6.1) :

gh.cpp:11:5: error: could not convert '{1, "ONE", 2, "TWO"}' from '<brace-enclosed initializer list>' to 'const std::map<int, std::basic_string<char> >'

I know how to fix it :

  const std::map< int, std::string > mapping = {
      {1, "ONE"},
      {2, "TWO"},
    };

but why the compilation fails in the top example?

Answer 1

Because the map is a non-aggregate, and contains non-aggregate elements (std::pair<key_type, mapped_type>), so it requires an initializer-list full of initializer-lists, one for each pair.

std::pair<int,int> p0{ 1,2 }; // single pair
std::map<int, int> m { { 1,2 } }; // map with one element
std::map<int, int> m { { 1,2 }, { 3,4} }; // map with two elements

Bear in mind that the rules for brace elision apply to aggregates, so they do not apply here.