How to use make_pair to create pair of id and struct (object)?

Question

I was trying to create a pair of id and object like this:

 #include <iostream>
  #include <utility>
  #include <functional>

  struct num{

      double x;
      double y;

  };


  int main(){

      auto tmp = std::make_pair(1, {1.0, 2.0});

 }

I get error as error: no matching function for call to 'make_pair(int, <brace-enclosed initializer list>)'

Is there a proper way to create pair of id and object?

How is the compiler supposed to guess that {1.0, 2.0} should be a num? — NathanOliver
– NathanOliver, Commented Dec 21, 2016 at 19:40
What type would you expect to be deduced automatically for auto tmp?? Your struct is completely unrelated (not introduced at all). — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Dec 21, 2016 at 19:41

Humam Helfawi · Accepted Answer · 2016-12-21 21:00:27Z

8

No, This is how you should create your pair:

auto tmp = std::make_pair(1, num{1.0, 2.0});

Or alternatively (as @StoryTeller mentioned):

std::pair<int,num> tmp {1, {1.0, 2.0}};

Now, in both cases, the compiler has a clue that {1.0, 2.0} is meant to be an initializer for the num.

edited Dec 21, 2016 at 21:00

answered Dec 21, 2016 at 19:40

Humam Helfawi

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6 Comments

StoryTeller - Unslander Monica Over a year ago

Personally, if naming the types is already a must, I prefer to shorten the line by simply writing std::pair<int,num> temp{1, {1.0, 2.0}};

Praetorian Over a year ago

Never specify template arguments when using make_pair, just use the pair constructor in that case

solti Over a year ago

@Praetorian may I ask why ?

Praetorian Over a year ago

@solti For one, it's less typing. And then there are cases like this where things won't compile int i = 1; auto tmp = std::make_pair<int,num>(i, {1.0, 2.0}); because of how make_pair is defined. Quoting STL - Don't help the compiler

Humam Helfawi Over a year ago

@Praetorian Thank you very much for mentioning the video

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