Using Boost adaptors with C++11 lambdas

Question

I tried to compile this code:

#include 
#include 
#include 

int main() {
             


        

Answer 1

http://smellegantcode.wordpress.com/2011/10/31/linq-to-c-or-something-much-better/

But you can use this, that works well.

#include <boost/range/adaptors.hpp>
#include <boost/range/algorithm.hpp>
#include <vector>
#include <functional>

int main() {
    std::vector<int> v{
        1,5,4,2,8,5,3,7,9
    };
    std::function<int(int)> func = [](int i) { return -i; };
    std::cout << *boost::min_element(v | boost::adaptors::transformed(
    func)) << std::endl;
    return 0;
}

http://liveworkspace.org/code/b78b3f7d05049515ac207e0c12054c70

#define BOOST_RESULT_OF_USE_DECLTYPE works fine in VS2012 for example.

Answer 2

With C++17 feature class template argument deduction you can wrap with std::function simplier, like this:

*boost::min_element(
    v | boost::adaptors::transformed(std::function(
        [](int i) { return -i; })));

Answer 3

This is covered at http://boost.2283326.n4.nabble.com/range-cannot-use-lambda-predicate-in-adaptor-with-certain-algorithms-td3560157.html and https://svn.boost.org/trac/boost/ticket/4189 - the problem is that some algorithms expect to be able to copy-construct (and default-construct, and copy-assign) their predicate, which can't be done with a lambda.

The workaround is to wrap the lambda in a std::function:

*boost::min_element(
    v | boost::adaptors::transformed(std::function<int(int)>(
        [](int i) { return -i; })));

I've asked (at Inferring the call signature of a lambda or arbitrary callable for "make_function") for a way to write a make_function such that one can just write:

*boost::min_element(
    v | boost::adaptors::transformed(make_function(
        [](int i) { return -i; })));

Answer 4

You can turn a non-capturing lambda into a function pointer by putting a "+" in front of it.

std::vector<int> v{1,5,4,2,8,5,3,7,9};
std::cout << *boost::min_element(v | 
    boost::adaptors::transformed(+[](int i) 
    {
        return -i; 
    })) << std::endl;