How can I use a boost function to transform the parameter types?

Question

I want to create a boost function object of the following signature:

void (int, boost::uuid);

However, I would like to bind it to a function of the following form:

void (SomeType, boost::uuid)

Where the SomeType argument comes from another function call, so that if I were to call it straight out it would look like:

SomeType myOtherFunction(int);//Prototype 
... 
myFunction(myOtherFunction(int), myUUID);

In other words, I want the top level function object to completely hide the concept of SomeType and the call to myOtherFunction from the user. Is there a way to do this with one or more boost::function objects created with boost::bind calls?

score 3 · Accepted Answer · answered Sep 24 '14 at 01:02

Functional composition: Live On Coliru

#include <boost/uuid/uuid.hpp>

struct SomeType {};
SomeType myOtherFunction(int) { return SomeType(); }
void foo(SomeType, boost::uuids::uuid) {}

#include <boost/bind.hpp>
#include <boost/function.hpp>

int main()
{
    boost::function<void(int, boost::uuids::uuid)> composed;

    composed = boost::bind(foo, boost::bind(myOtherFunction, _1), _2);
}

Anyways, in c++11 you'd write [](int i, uuid u) { return foo(myOtherFunction(i), u); } of course

How can I use a boost function to transform the parameter types?

1 Answers1