Thank you Steven. While I do not have boost installed on this computer, the code below compiles and runs as I would expect. Upon each execution, the value of seed changes. I expect boost::mt19937 rng(seed); will work. And now I have a very good random seed thanks to your advice. Your friend, Ed #include <iostream> #include <windows.h> #include <wincrypt.h> int main() { HCRYPTPROV hProvider = 0; BYTE randomBytes[8]; CryptAcquireContext(&hProvider, 0, 0, PROV_RSA_FULL, CRYPT_VERIFYCONTEXT); CryptGenRandom(hProvider, sizeof(randomBytes), randomBytes); CryptReleaseContext(hProvider, 0); uint64_t seed; memcpy(&seed, &randomBytes[0], sizeof(seed)); std::cout << seed << std::endl; return 0; } On Wed, Feb 10, 2010 at 11:58 PM, Steven Watanabe <watanabesj@gmail.com>wrote:
AMDG
e m wrote:
I have a C++ program that uses /dev/urandom on Unix systems to seed boost::mt19937. I'm porting the program to Windows. I can get good, random seed data on Windows similar to random seed data produced from /dev/urandom on Unix systems like so:
HCRYPTPROV hProvider = 0; BYTE randomBytes[8]; CryptAcquireContext(&hProvider, 0, 0, PROV_RSA_FULL, CRYPT_VERIFYCONTEXT); CryptGenRandom(hProvider, sizeof(randomBytes), randomBytes); CryptReleaseContext(hProvider, 0);
However, I'm running into trouble attempting to pass these randomBytes to boost::mt19937. On Unix I do something similar to this:
uint64_t GetSeed() { uint64_t seed; std::ifstream urandom; urandom.open("/dev/urandom"); urandom.read(reinterpret_cast<char*> (&seed), sizeof(seed)); urandom.close(); return seed; }
boost::mt19937 rng(GetSeed());
That works OK on Unix I think. Could anyone suggest how to get the random data from the Windows CryptoAPI passed to boost::mt19937 as seed data? I don't normally program Windows systems, so this has been a bit of a challenge for me.
Try
uint64_t seed; memcpy(&seed, &randomBytes[0], sizeof(seed));
In Christ, Steven Watanabe
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