CODEWITHSUNDEEP
c++c++11initializationc++14unordered-map
ilya1725
ilya1725

Reputation: 4938

Initialize unordered_map in the initializer list

I'm trying to find a solution to what may be a very trivial problem. I would like to initialize my const unordered_map in the class initializer list. However I'm yet to find the syntax that the compiler (GCC 6.2.0) will accept. A code link is here.

#include <unordered_map>

class test {
 public:
    test()
      : map_({23, 1345}, {43, -8745}) {}

 private:
   const std::unordered_map<long, long> map_;
 };

Error:

main.cpp: In constructor 'test::test()':
main.cpp:6:36: error: no matching function for call to 'std::unordered_map<long int, long int>::unordered_map(<brace-enclosed initializer list>, <brace-enclosed initializer list>)'
  : map_({23, 1345}, {43, -8745}) {}
                                ^

Are the complex constants not allowed to be initialized in the initializer list? Or the syntax has to be different?

Upvotes: 9

Views: 30840

Answers (2)

Guan Wang
Guan Wang

Reputation: 31

Use curly braces instead of the parentheses because if you use parentheses it's calling the constructor that best matches your arguments instead of the overloaded constructor with a parameter of type initializer_list.

Using parentheses and curly braces have the same effect until there's an overloaded constructor taking initializer_list type as a parameter. Then when you use curly braces, the compiler is going to bend over backwards to try to call that overloaded constructor.

for example:

Foo( 3 ) is calling Foo( int x ) constructor;
Foo{ 3 } is calling Foo( initializer_list<int> x ) constructor;

but if there's no Foo( initializer_list<int> x ) constructor

then Foo( 3 ) and Foo{ 3 } are both calling Foo( int x ) constructor.

Upvotes: 3

Vlad from Moscow
Vlad from Moscow

Reputation: 310990

Use braces instead of the parentheses

class test {
 public:
    test()
      : map_{{23, 1345}, {43, -8745}} {}

 private:
   const std::unordered_map<long, long> map_;
 };

Upvotes: 16

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