[Boost-users] boost::lambda and math functions

[Peter Foelsche]

I've the following template function:

template<typename T>
T exp2(const T T0_d)
{ return exp(0.5*T0_d) + 2*exp(T0_d);
}

I want to define this as a lambda function in place (but don't know how).

It will be used with T=double and T=MyAutomaticDerivativeClass.
I've already defined the return type of arithmetic operators applied to
MyAutomaticDerivativeClass:

namespace boost {
namespace lambda {

template<class Act, typename X, typename Y>
struct plain_return_type_2<arithmetic_action<Act>, X, Y>
{ typedef ... type;
};

}
}

What else is missing?

Thanks
Peter

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[Roman Perepelitsa]

2009/11/24 Peter Foelsche <peter_f...@agilent.com>

I've the following template function:


template<typename T>
T exp2(const T T0_d)
{ return exp(0.5*T0_d) + 2*exp(T0_d);
}


I want to define this as a lambda function in place (but don't know how).

It will be used with T=double and T=MyAutomaticDerivativeClass.

#include <cmath>

#include <boost/lambda/lambda.hpp>

#include <boost/lambda/bind.hpp>

int main() {

  using namespace boost::lambda;

  double res = (bind(exp, 0.5 * _1) + 2 * bind(exp, _1))(0.5);

}

If exp is overloaded (and I guess it is, because you are going to pass both double and MyAutomaticDerivativeClass as arguments to exp), you'll need to wrap it in a polymorphic function object.

struct Exp {

  template <class Args> 

  struct sig : boost::tuples::element<1, Args> {};

  template <class T>

  T operator()(T arg) const {

    return exp(arg);

  }

};

int main() {

  using namespace boost::lambda;

  double res = (bind(Exp(), 0.5 * _1) + 2 * bind(Exp(), _1))(0.5);

}

By the way, if you have a choice, I recommend using phoenix instead of lambda. It can do everything lambda can do and even more.

Roman Perepelitsa.