You need to pass the type of the lambda as a template argument, not the lambda itself. What you want is this:
auto mycomp = [](const int&a, const int& b) { return a < b; };
std::map<int, int, decltype(mycomp)> test(mycomp);
Although in fact, since your lambda has no captures, it can actually be stored in a function pointer, so alternatively, you could do this:
std::map<int, int, bool(*)(const int&,const int&)>
test([](const int&a, const int& b) { return a < b; });
Though I find the first much more readable. Although using the function pointer type is more versatile. i.e. It can accept any function pointer or non-capturing lambda that matches that signature. But if you change your lambda to be capturing, it will not work. For a more versatile version, you could use std::function
, i.e:
std::map<int, int, std::function<bool(const int&, const int&)>>
That will work with any function, lambda(capturing or not) or function object, as long as the signature matches.
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