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Simple STL Question

Started by bakery2k1 Apr 10, 2005 at 3:41 PM 1 replies 800+ views

Original Post

Apr 10, 2005 at 3:41 PM

I am working through Nicolai Josuttis' book on the C++ Standard Library, and am attempting to implement a simple functor. I have:


template<class T>
class Sum
{
public:
	Sum() : sum(0)
	{}

	void operator()(const T & elem)
	{ sum+=elem; }

	T sum;
};

int main()
{
	std::vector<int> coll;

	for(int i=0; i<5; ++i)
		coll.push_back(i);

	Sum<int> sum;

	for(std::vector<int>::const_iterator ci = coll.begin(); ci != coll.end(); ++ci)
		sum(*ci);

	std::cout << sum.sum << std::endl;
	
	return 0;
}

Does anyone know why this works, but replacing the for loop with:


	std::for_each(coll.begin(), coll.end(), sum);

means that the result is zero every time?

Fruny

Apr 10, 2005 at 3:43 PM

Because the functor is passed by value. You've got to grab the return value from std::for_each if you want the post-iteration functor:

sum = std::for_each(coll.begin(), coll.end(), sum);

me22

Apr 10, 2005 at 4:27 PM

I know it's not your question, but I'd like to point you to the accumulate function:

std::cout << std::accumulate( coll.begin(), coll.end(), static_cast<col1::value_type>(0) ) << std::endl;

Note that the return type is the same as the type of the third parameter, so if you say accumulate( my_floats.begin(), my_floats.end(), 0 ) you'll get very annoyed at the STL ( hence the explicit cast on 0 ).

Also, I don't think there's anything in the specs that prevents for_each from making a copy internally of the functor, so there's no way to make that idea work. ( Short of having Sum contain a reference to the variable to increment which would be set by the constructor, which is alot more work than just using accumulate... )

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