srand() + rand() with local scope

Question

I have a function that calls srand and rand like this:

void foo() { int seed = some_operation(); std::srand(seed); int value = std::rand(); // Do something with random value }

However, I don't want to change the global state of rand. Whats the easiest way to get a random number then?

Requirements:

random number must be deterministic based on seed
C++11 is fine
foo should be thread safe
the global state of rand should not be modified

Edit:

There is a stackoverflow question asking how to generate random numbers. The accepted answer however shows how to generate truly unique random numbers, using a slow std::random_device. I just needed a simple generator using a fixed seed instead.

possible duplicate of Best way to generate random numbers using c++11 random library — eerorika
– eerorika, Commented May 27, 2015 at 14:09
Also Random number generation in C++11 , how to generate , how do they work? — AliciaBytes
– AliciaBytes, Commented May 27, 2015 at 14:10

Mike Seymour · Accepted Answer · 2015-05-27 14:16:06Z

C++11 is fine

Then use the new pseudorandom number library:

#include <random> int foo() { int seed = some_operation(); std::minstd_rand rand(seed); int value = rand(); // Do something with random value }

minstd_rand is a simple linear congruential engine, similar to that typically used by std::rand(), but with its state encapsulated in a class. Other engines are available, if you need a higher quality pseudorandom sequence. The Mersenne Twister, std::mt19937, is usually a good choice if you don't have specific requirements.

@Brahim: To protect what? The engine is a local variable, so can only be accessed by the thread that's calling the function.
You are right. I just thought that the library functon's thread-safety may be undefined.

Freyja · Accepted Answer · 2015-05-27 14:14:17Z

<random> defines a number of PRNG classes that you can use, which do not use global states.

For example, using a default Mersenne Twister, std::mt19937:

#include <iostream> #include <random> int main() { int seed = 1234; std::mt19937 rng(seed); std::cout << "Random number: " << rng() << std::endl; }

Brent Bradburn · Accepted Answer · 2015-05-27 14:14:33Z

1

One approach is to provide your own implementation, with its own seed data. A template for doing that is provided in the manpage for rand().

static unsigned long next = 1; /* RAND_MAX assumed to be 32767 */ int myrand(void) { next = next * 1103515245 + 12345; return((unsigned)(next/65536) % 32768); } void mysrand(unsigned seed) { next = seed; }

answered May 27, 2015 at 14:14

Brent Bradburn

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3 Comments

DanielKO Over a year ago

He asked for a solution with no global state, and you post an answer with global state.

Brent Bradburn Over a year ago

@DanielKO: Actually, he said he didn't want to change the global state of rand(), which this does not. I get your point, however, and was kind of expecting that someone could modify this to manage the state in a way that suited their purposes -- for example if they wanted one seed per thread, that would require custom logic which isn't addressed in any of the answers.

Brent Bradburn Over a year ago

Note: The above code was copied verbatim from the man page. It is really only an example of how you might generate a predictable random number sequence in a very portable way (and with custom state data). Approaches using <random> are likely better if you are targeting C++11 specifically. The code, as written, is not thread-safe. To achieve thread-safety you need to manage the state variable next in a way that matches your thread safety strategy.

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