需要从枚举（类）到 std::binary_function 的 map

Question

I have this enum (class)

enum class conditional_operator
{
    plus_op,
    or_op,
    not_op
}

And I'd like a std::map that represents these mappings:

std::map<conditional_operator, std::binary_function<bool,bool,bool>> conditional_map = 
        { { conditional_operator::plus_op, std::logical_and<bool> },
          { conditional_operator::or_op,   std::logical_or<bool>},
          { conditional_operator::not_op,  std::binary_negate<bool>} // this seems fishy too, binary_negate is not really what I want :(

Apart from the fact that this doesn't compile:

error: expected primary-expression before '}' token

error: expected primary-expression before '}' token

error: expected primary-expression before '}' token

for each of the three lines, how should I do this? I think a logical_not with a second dummy argument would work, once I get this to compile of course...

EDIT: Could I use lambda's for this?

Answer 1

You really want std::function<bool(bool, bool)> , not std::binary_function<bool, bool, bool> . That only exists for typedefs and stuff in C++03. Secondly, I'd just use a lambda- they're short enough and much clearer. The std::logical_and and stuff only exists for C++03 function object creation, and I'd use a lambda over them any day.

std::map<conditional_operator, std::function<bool(bool,bool)>> conditional_map = 
{ 
    { conditional_operator::plus_op, [](bool a, bool b) { return a && b; } },
    { conditional_operator::or_op,   [](bool a, bool b) { return a || b; } },
    { conditional_operator::not_op,  [](bool a, bool b) { return !a; } }
};

Wait- what exact operator are you referring to with not? Because that's unary, as far as I know.

Answer 2

@DeadMG's answer is spot-on but if you insist on using the predefined function objects, you need to instantiate them. At the moment you're just passing their type names.

That is, you need to write std::logical_***<bool>() instead of just std::logical_***<bool> .