Initialize std::map, using default-constructed values

Question

If I have a std::map<int, char>, it's nice and easy to initialise:

std::map<int, char> myMap = 
{
  { 1, 'a' },
  { 2, 'b' }
};

If my value-type has a default constructor, I can initialise it like so:

struct Foo
{
  char x;
};

std::map<int, Foo> myMap =
{
  { 1, Foo() },
  { 2, Foo() }
};

But suppose my value-type has a default constructor, but is unwieldy to type:

std::map<int, yourNamespace::subLibrary::moreLevels::justBecause::Thing> myMap = 
{
  // What goes here? I don't want to type
  // yourNamespace::subLibrary::moreLevels::justBecause::Thing again.
};

Short of a using declaration to make the value-type more keyboard-friendly, is there some way to initialise the map with a set of keys, all of which use a default-constructed yourNamespace::subLibrary::moreLevels::justBecause::Thing? How can I signal that I want the mapped value to be default-constructed of its type?

François Andrieux · Accepted Answer · 2017-01-16 16:35:18Z

4

You can use value initialization to solve this problem.

struct Foo
{
    char x;
};

std::map<int, Foo> myMap =
{
    { 1, {} },
    { 2, {} }
};

answered Jan 16, 2017 at 16:35

François Andrieux

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5 Comments

Chowlett Over a year ago

I thought I could, and I tried it, and it didn't work... which is why I came here. Further prodding shows it works in general, but not if the value-type is not copy-constructible. Do you know why not?

François Andrieux Over a year ago

@Chowlett Regarding your follow-up question, I will refer you to this answer.

Chowlett Over a year ago

Ah. That would explain it. Am I therefore completely stuffed for initialising in this fashion? I note that even myMap = { { 1, Foo() } }; fails, for presumably exactly the same reason.

François Andrieux Over a year ago

@Chowlett If you don't have a copy constructor, you will have a bad time with initializer lists. Consider asking a separate question on how to work around this problem. Include what you've tried so far.

Sandburg Over a year ago

additional info: Since C++11

Richard Assal · Accepted Answer · 2017-01-16 16:44:26Z

0

You could make a function that returns an instance of "yourNamespace::subLibrary::moreLevels::justBecause::Thing".

answered Jan 16, 2017 at 16:44

Richard Assal

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