[std::set] lambda as less predicate

[morando]

Why this doesn't work?

auto less_fun = [](const Pair& l, const Pair& r)->bool{
			return l.second < r.second;
		};
		std::set<Pair, less_fun> sn;

output error:

error C2923: 'std::set' : 'less_fun' is not a valid template type argument for parameter '_Pr'

I am using VS 2012.

Thank you in advance.

[Duthomhas]

'less_fun' is an object, an instance of a type of thing. You have avoided having to know its exact type by using the auto keyword.

The std::set takes two types as argument. 'Pair' is a type, but 'less_fun' is not -- it is an object.

Use decltype to get its type:

std::set<Pair, decltype(less_fun)> sn;


Personally, I prefer more_fun. :-)

[morando]

I tried that before, but still doesnt work, error output:

error C3497: you cannot construct an instance of a lambda

Thank you for your time.

[JLBorges]

A lambda expression is always evaluated; it can not not appear in an unevaluated context. And it has no default constructor.

#include <set>
#include <functional>

int main()
{
    // sizeof( [] ( int v ) { return v ; } ) ; // error: expression is in unevaluated context
    // using x = decltype( [] ( int v ) { return v ; } ) ; // error: expression is in unevaluated context)

    auto lambda = [] ( int v ) { return v ; } ;

    sizeof(lambda) ; // ok, expression has been evaluated
    using y = decltype( lambda ) ; // ok, expression has been evaluated

    // y yy ; // error: a closure type has no default constructor (deleted)

    auto fun = [] ( int x, int y ) { return x%100 < y%100 ; } ;

    // std::set< int, decltype(fun) > set_one ; // error: no default constructor
    std::set< int, decltype(fun) > set_one(fun) ; // fine

    // recommended 
    std::set< int, std::function< bool(int,int) > > set_two(
                                            [] ( int x, int y ) { return x>y ; } ) ; // fine
}

[Peter87]

You need to pass less_fun as argument to the constructor.
std::set<Pair, decltype(less_fun)> sn(less_fun);

[morando]

Thank you all.

Works with std::function but still not with decltype:

@JLBorges

std::set< int, decltype(fun) > set_one(fun) ; // fine 

Got same error as before.

Thank you for your time.

EDIT: Works with Mingw 4.6.2 but not with VS2012

[vlad from moscow]

@JLBorges
sizeof(lambda) ; // ok, expression has been evaluated

The sizeof operator does not evaluate expressions.

[JLBorges]

> The sizeof operator does not evaluate expressions.

Yes, it does not. Which is why I used the the present perfect non-progressive tense "has been evaluated" (to indicate a past action that has an influence on the present).

Perhaps it would have made things clearer if the past perfect non-progressive tense was used instead, and if it was clarified that the expression being discussed is the lambda expression. ("the lambda expression had been evaluated")