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Home/ Questions/Q 975741
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T03:38:12+00:00 2026-05-16T03:38:12+00:00

Right now, I have this code: bool isAnyTrue() { for(std::list< boost::shared_ptr<Foo> >::iterator i =

  • 0

Right now, I have this code:

bool isAnyTrue() {
    for(std::list< boost::shared_ptr<Foo> >::iterator i = mylist.begin(); i != mylist.end(); ++i) {
        if( (*i)->isTrue() )
            return true;
    }

    return false;
}

I have used Boost here and then but I couldn’t really remember any simple way to write it somewhat like I would maybe write it in Python, e.g.:

def isAnyTrue():
    return any(o.isTrue() for o in mylist)

Is there any construct in STL/Boost to write it more or less like this?

Or maybe an equivalent to this Python Code:

def isAnyTrue():
    return any(map(mylist, lambda o: o.isTrue()))

Mostly I am wondering if there is any existing any (and all) equivalent in Boost / STL yet. Or why there is not (because it seems quite useful and I use it quite often in Python).

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  1. Editorial Team
    Editorial Team
    2026-05-16T03:38:13+00:00Added an answer on May 16, 2026 at 3:38 am

    C++ does not (yet) have a foreach construct. You have to write that yourself/

    That said, you can use the std::find_if algorithm here:

    bool isAnyTrue()
    {
        return std::find_if(mylist.begin(), mylist.end(), std::mem_fun(&Foo::isTrue))
               != mylist.end();
    }
    

    Also, you should probably be using std::vector or std::deque rather than std::list.

    EDIT: sth has just informed me that this won’t actually compile because your list contains shared_ptr instead of the actual objects… because of that, you’re going to need to write your own functor, or rely on boost:

    //#include <boost/ptr_container/indirect_fun.hpp>
    
    bool isAnyTrue()
    {
        return std::find_if(mylist.begin(), mylist.end(), 
               boost::make_indirect_fun(std::mem_fun(&Foo::isTrue))) != mylist.end();
    }
    

    Note, I haven’t tested this second solution.

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