View Problem
Generate a repeatable random number sequence
Initialise a random number generator with a seed and generate five decimal values. Reset the seed and produce the same values.
perl
cpp C++/CLI .NET 2.0
cpp
csharp
fsharp
Submit a new solution for perl, cpp, csharp, or fsharp
There are 13 other solutions in additional languages (clojure, erlang, fantom, go ...)
srand(12345);
@list1 = map(int(rand(100)+1), (1..5));
srand(12345);
@list2 = map(int(rand(100)+1), (1..5));
print join(', ', @list1) . "\n";
print join(', ', @list2) . "\n";
@list1 = map(int(rand(100)+1), (1..5));
srand(12345);
@list2 = map(int(rand(100)+1), (1..5));
print join(', ', @list1) . "\n";
print join(', ', @list2) . "\n";
void printAction(int i) { Console::Write("{0} ", i); }
array<int>^ genFillRand(array<int>^ arr, Random^ rnd, int lb, int ub)
{
for (int i = 0; i < arr->Length; ++i) arr[i] = rnd->Next(lb, ub + 1); return arr;
}
int main()
{
array<int>^ arr1 = genFillRand(gcnew array<int>(5), gcnew Random(12345), 100, 200);
array<int>^ arr2 = genFillRand(gcnew array<int>(5), gcnew Random(12345), 100, 200);
Action<int>^ print = gcnew Action<int>(printAction);
Array::ForEach<int>(arr1, print); Console::WriteLine();
Array::ForEach<int>(arr2, print); Console::WriteLine();
}
array<int>^ genFillRand(array<int>^ arr, Random^ rnd, int lb, int ub)
{
for (int i = 0; i < arr->Length; ++i) arr[i] = rnd->Next(lb, ub + 1); return arr;
}
int main()
{
array<int>^ arr1 = genFillRand(gcnew array<int>(5), gcnew Random(12345), 100, 200);
array<int>^ arr2 = genFillRand(gcnew array<int>(5), gcnew Random(12345), 100, 200);
Action<int>^ print = gcnew Action<int>(printAction);
Array::ForEach<int>(arr1, print); Console::WriteLine();
Array::ForEach<int>(arr2, print); Console::WriteLine();
}
typedef boost::uniform_int<> Distribution;
typedef boost::mt19937 RNG;
Distribution distribution(100, 200);
RNG rng;
boost::variate_generator<RNG&, Distribution> generator(rng, distribution);
rng.seed(42L);
std::generate_n(std::ostream_iterator<unsigned>(std::cout, " "), 5, generator);
rng.seed(42L);
std::cout << std::endl;
std::generate_n(std::ostream_iterator<unsigned>(std::cout, " "), 5, generator);
typedef boost::mt19937 RNG;
Distribution distribution(100, 200);
RNG rng;
boost::variate_generator<RNG&, Distribution> generator(rng, distribution);
rng.seed(42L);
std::generate_n(std::ostream_iterator<unsigned>(std::cout, " "), 5, generator);
rng.seed(42L);
std::cout << std::endl;
std::generate_n(std::ostream_iterator<unsigned>(std::cout, " "), 5, generator);
using System;
public class RepeatableRandom {
public static void Main() {
var r = new Random(12); // seed is 12
for (int i = 0; i < 5; i++)
Console.WriteLine(r.Next());
r = new Random(12);
for (int i = 0; i < 5; i++)
Console.WriteLine(r.Next());
}
}
public class RepeatableRandom {
public static void Main() {
var r = new Random(12); // seed is 12
for (int i = 0; i < 5; i++)
Console.WriteLine(r.Next());
r = new Random(12);
for (int i = 0; i < 5; i++)
Console.WriteLine(r.Next());
}
}
let (seed, lb, ub) = (12345, 100, 200)
let mutable rnd = new Random(seed)
for i = 1 to 5 do printf "%d " (rnd.Next(lb, ub + 1)) done ; printfn ""
rnd <- new Random(seed)
for i = 1 to 5 do printf "%d " (rnd.Next(lb, ub + 1)) done ; printfn ""
let mutable rnd = new Random(seed)
for i = 1 to 5 do printf "%d " (rnd.Next(lb, ub + 1)) done ; printfn ""
rnd <- new Random(seed)
for i = 1 to 5 do printf "%d " (rnd.Next(lb, ub + 1)) done ; printfn ""
Submit a new solution for perl, cpp, csharp, or fsharp
There are 13 other solutions in additional languages (clojure, erlang, fantom, go ...)
