Is the library in c++11 portable? I have avoided rand() because I heard it wasn't portable.
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How do you define "portable"?
If by "portable", you mean "will produce binary identical sequences of random numbers given the same input", then yes, rand isn't portable. And yes, the C++ random generators are portable (most of them. Not std::default_random_engine or std::random_device), because they implement specific algorithms. rand is allowed to be anything, as long as it's not entirely unlike a random number generator.
That being said, as @PeteBecker pointed out, the distributions themselves are not so well-defined. So while std::mt19937 will produce the same sequence of values for a given seed, different std::uniform_int_distributions can give different values for the same input sequence and range.
Of course, if you need consistency, you can always define your own distribution.
Nicol Bolas
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3Yes that is what I meant by portable. – Xavier Feb 12 '13 at 20:13
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You can find the rationale for the distributions being implementation defined in [my answer here](http://stackoverflow.com/a/24554535/1708801). – Shafik Yaghmour Jul 03 '14 at 14:19
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The random number engines described in <random> have explicit requirements for their algorithms to ensure portability. The distributions do not.
Pete Becker
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You can generate "identical sequences of random numbers given the same input" (from @Nicol Bolas) with std::mt19937 (Mersenne Twister) for example. You definitely couldn't do that with rand() which was quite annoying.
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