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Function Pointer Adapters

The header functional.hpp provides enhanced versions of both the function pointer adapters from the C++ Standard Library (§20.3.7):

As well as the corresponding helper function template:

However, you should not need to use the adapters in conjunction with the adapters in this library due to our use of function object traits. You will however need to use them if your implementation fails to work properly with our traits classes (due to lack if partial specialisation), or if you wish to use a function object adapter from a third party.


If you need to use these adapters, usage is identical to the standard function pointer adapters. For example,

bool bad(std::string foo) { ... }
std::vector<std::string> c;
std::vector<std::string>::iterator it
     = std::find_if(c.begin(), c.end(), std::not1(boost::ptr_fun(bad)));

Note however that this library contains enhanced negators that support function object traits, so the line above could equally be written

     = std::find_if(c.begin(), c.end(), boost::not1(bad));

Argument Types

The standard defines pointer_to_unary_function like this (§20.3.8 ¶2):

template <class Arg, class Result>
class pointer_to_unary_function : public unary_function<Arg, Result> {
  explicit pointer_to_unary_function(Result (* f)(Arg));
  Result operator()(Arg x) const;

Note that the argument to operator() is exactly the same type as the argument to the wrapped function. If this is a value type, the argument will be passed by value and copied twice. pointer_to_binary_function has a similar problem.

However, if we were to try and eliminate this inefficiency by instead declaring the argument as const Arg&, then if Arg were a reference type, we would have a reference to a reference, which is currently illegal (but see C++ core language issue number 106)

So the way in which we want to declare the argument for operator() depends on whether or not the wrapped function's argument is a reference. If it is a reference, we want to declare it simply as Arg; if it is a value we want to declare it as const Arg&.

The Boost call_traits class template contains a param_type typedef, which uses partial specialisation to make precisely this decision. By declaring the operator() as

Result operator()(typename call_traits<Arg>::param_type x) const

we achieve the desired result - we improve efficiency without generating references to references.


The call traits template used to realise this improvement relies on partial specialisation, so this improvement is only available on compilers that support that feature. With other compilers, the argument passed to the function will always be passed by reference, thus generating the possibility of references to references.

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Revised 02 December, 2006

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