CODEWITHSUNDEEP
c++c++17priority-queue
mzlo
mzlo

Reputation: 353

C++ : How to define map of priority queues using lambda comparator?

You can define priority queue with lambda as

auto cmp = [] (const MyClass& a, const MyClass& b) -> bool{return a.v2 > b.v2};
priority_queue<MyClass, std::vector<MyClass>, decltype(cmp)> pq(cms);

When I use function object(custom comparison) to define map of priority queue, there is no error.

   unordered_map<int, priority_queue<int, std::vector<int>, greater<int>> my_map;

But when I use same technique(custom comparison using lambda) to define map of priority queue I get error.

unordered_map<MyKeyClass, priority_queue<MyClass, std::vector<MyClass>, decltype(cmp)>> my_map;

Error

/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/queue:447:16: error: no matching constructor for initialization of 'std::__1::priority_queue<int, std::__1::vector<int, std::__1::allocator<int>
      >, (lambda at avg5.cpp:13:24)>::value_compare' (aka '(lambda at avg5.cpp:13:24)')
        : c(), comp() {}
               ^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/tuple:1401:7: note: in instantiation of member function 'std::__1::priority_queue<int, std::__1::vector<int, std::__1::allocator<int> >,
      (lambda at avg5.cpp:13:24)>::priority_queue' requested here
      second(_VSTD::forward<_Args2>(_VSTD::get<_I2>(__second_args))...)
      ^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/utility:507:11: note: in instantiation of function template specialization 'std::__1::pair<const int, std::__1::priority_queue<int,
      std::__1::vector<int, std::__1::allocator<int> >, (lambda at avg5.cpp:13:24)> >::pair<const int &, 0>' requested here
        : pair(__pc, __first_args, __second_args,
          ^

How to define and populate map of priority queues with custom comparison using lambda ?

Upvotes: 2

Views: 399

Answers (1)

J&#233;r&#244;me Richard
J&#233;r&#244;me Richard

Reputation: 50308

You should not manipulate the type of lambdas. You should use std::function in your case. Here is what the standard say about lambda types:

The lambda expression is a prvalue expression of unique unnamed non-union non-aggregate class type, known as closure type, which is declared (for the purposes of ADL) in the smallest block scope, class scope, or namespace scope that contains the lambda expression. [...]
Lambda-expressions are not allowed in unevaluated expressions, template arguments, alias declarations, typedef declarations, and anywhere in a function (or function template) declaration except the function body and the function's default arguments.

You can find more information about this here and here.

Upvotes: 1

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