Question

How do I create a lambda function using boost or the stl to match the boost::function parameter expected by F in the third snippet of code in main?

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

void F(int a, boost::function<bool(int)> f) {
    std::cout << "a = " << a << " f(a) = " << f(a) << std::endl;
}

bool G(int x) {
    return x == 0;
}

int main(int arg, char** argv) {
    // C++0x
    F(123, [](int i) { return i==0; } );

    // Using seperate function
    F(0, &G);

    // How can I do it in place without C++0x
    F(123, /* create a lambda here to match */);
}

I can’t use C++0x and would like to avoid creating several separate functions. I can use something other that boost::function if that helps, my priority is creating the lambda succinctly.

Answer 1

#include <functional>    // STL
#include <boost/lambda/lambda.hpp>   // Boost.Lambda
#include <boost/spirit/include/phoenix_core.hpp>     // Boost.Pheonix
#include <boost/spirit/include/phoenix_operator.hpp> // Boost.Pheonix also

...

// Use STL bind without lambdas
F(0, std::bind2nd(std::equal_to<int>(), 0));
F(123, std::bind2nd(std::equal_to<int>(), 0));

// Use Boost.Lambda (boost::lambda::_1 is the variable)
F(0, boost::lambda::_1 == 0);
F(123, boost::lambda::_1 == 0);

// Use Boost.Phoenix
F(0, boost::phoenix::arg_names::arg1 == 0);
F(123, boost::phoenix::arg_names::arg1 == 0);

You may want to add some using namespace to simplify the code.

Boost.Lambda is strictly for defining functors inline with a C++-like syntax, while Boost.Phoenix is a functional-programming language built on top of C++ abusing (☺) its syntax and compile-time computation capability. Boost.Phoenix is much more powerful than Boost.Lambda, but the former also takes much more time to compile.