How to make sure that std::random_shuffle always produces a different result?

问题

Is there some function, similar to srand(), that I need to call to make sure that std::random_shuffle() always produces different results? i.e. if I call it several times with the same data, I want the order to be different every time. How can I make sure of that?

回答1:

std::random_shuffle has two forms. One that takes 2 arguments (begin/end iterators), and one that takes 3 (begin/end iterator and a random generator).

The first form uses std::rand(), so you would use std::srand() to seed it's random number generator. You can also use the 3-argument version and provide the RNG yourself.

回答2:

std::random_shuffle has a template overload for specifying the RNG.

template <class RandomAccessIterator, class RandomNumberGenerator>
  void random_shuffle ( RandomAccessIterator first, RandomAccessIterator last,
                        RandomNumberGenerator& rand );

reference

回答3:

random_shuffle is deprecated since C++14 (removed in C++17) and replaced with shuffle (exists since C++11) http://en.cppreference.com/w/cpp/algorithm/random_shuffle

possible usage:

shuffle(items.begin(), items.end(), std::default_random_engine(std::random_device()()));

回答4:

Generally call srand(time(NULL)) before calling std::random_shuffle() would give you what you need, it give you different result each time you call std::random_shuffle(). It's because std::random_shuffle() internally calls rand() in many populate implementation(e.g VS2008 and gcc).

Of course you can supple a RNG yourself if you want to call the other overloaded std::random_shuffle with a extra parameter.

回答5:

I think you can give a random generator functor to std::random_shuffle, so you can be able to fully control the random number generation. Looking here, this functor takes the place of the RandomNumberGenerator template argument.