Re: [Boost-users] Boost Lambda Newbie question

On Nov 27, 2006, at 1:14 AM, Haroon Khan wrote:

Hi, I'm experimenting with the boost lambda library, and I wrote the following code, to go through the vector of numbers and print out statements if the number is greater than 5 or less than 5

int a[]={1,2,3,4,5,6,7,8,9,10,10,9,8,4,7,6,5,4,3,3,1}; std::vector<int> v(a,a+sizeof(a)/sizeof(int)); std::for_each ( v.begin() , v.end() , (if_(_1<5)[ std::cout<

The boost::basic_format::operator % code gives an assertion since the object that is being passed is the placeholder object, and not the current value. Is there a correct way of doing this?

There is, but it is kind of complicated. Lambda Library gets kind of hairy with more complex lambda expressions: #include #include #include #include #include #include <iostream> using namespace boost::lambda; int main() { int a[]={1,2,3,4,5,6,7,8,9,10,10,9,8,4,7,6,5,4,3,3,1}; std::vector<int> v(a,a+sizeof(a)/sizeof(int)); std::for_each ( v.begin(), v.end(), (if_(_1<5)[ std::cout << retboost::format(bind(constructorboost::format(), "%d is less than 5\n") % _1) ].else_[ std::cout << retboost::format(bind(constructorboost::format(), "%d is greater than 5\n") % _1) ]) ); } The bind(constructorboost::format(), ...) is necessary, because you really need to construct a new format object at every iteration, otherwise the format library throws an exception saying you are trying to pass to many parameters to a format object, more that what is specfied in the format string. the retboost::format(...) is needed to tell lambda what the return type of format's operator% is. All in all, the above lambda is complex enough that writing a function object explicitly might be more justified... at least until we have built-in lambdas, if that ever happens. Cheers, Jaakko

Thanks, Haroon _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users