How to accumulate a two dimensional array with std::for_each?
Code:std::vector<std::vector<long> > v; std::for_each(v.begin(), v.end(), ????);
How to accumulate a two dimensional array with std::for_each?
Code:std::vector<std::vector<long> > v; std::for_each(v.begin(), v.end(), ????);
When you say accumulate, are you thinking of std::accumulate instead of std::for_each()?
Anyway, it is quite simple: just make the function or function object able to handle a std::vector<long> as the argument.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
You mean something like this?
I dont see the advantage anymore of just doing it the "old fashioned" way then with a loop. Was hoping for a better solutionCode:class Acc { public: Acc() : m_total(0) {} void operator()(std::vector<long> &v) { m_total += std::accumulate(v.begin(), v.end(), 0); } private: long m_total; };
No.
You need to use `std::accumulate' correctly and ignore `std::for_each'.
In other words, use a tool that was designed to solve the problem you want to solve.
Edit: Not that you can't do it with `std::for_each'; you just need to use one or the other.
Soma
Last edited by phantomotap; 05-11-2009 at 04:35 AM.
Im just asking if there is a one-liner to accumulate all items in a vector of vector of long. Doesnt matter if I use accumulate or for_each or whatever
??Code:vector<vector<long> > v; total = sum(v.begin(), v.end(), 0);
If you are so keen on doing it as a one-liner, you might read on nesting STL algorithms with Boost.Lambda.
Perhaps it just ain't worth it...
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With a struct it might look like this (a bit simpler than yours, and the struct could technically be a free function):
Code:#include <vector> #include <numeric> #include <iostream> using namespace std; struct subaccumulator { long operator()(long r, const vector<long>& v) const { return r + accumulate(v.begin(), v.end(), 0l); } }; int main() { vector<vector<long> > v(2, vector<long>(3, 4)); //2 by 3 all 4's cout << accumulate( v.begin(), v.end(), 0l, subaccumulator()) << '\n'; }
Last edited by anon; 05-11-2009 at 05:36 AM.
I might be wrong.
Quoted more than 1000 times (I hope).Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
"Im just asking if there is a one-liner to accumulate all items in a vector of vector of long."
Yes. Infinitely many...
Of course, that doesn't mean that you will not have to code some support routines yourself, but then that's the beauty of C++ generics. You will only have to code it once!
Soma
