C++ Return (uniform_int_distribution) from a function

Question

I made this random class so I can use random numbers in my game engine whenever I need, but I'm having a problem with the functions, How I'm I suppose to return R ? (please keep in mind that I'm new to coding in general)

#pragma once
#include <ctime>
#include <random>

using namespace std;

class Random
{
private:


public:
    Random() { default_random_engine RandomNumber(time(0));};

    int RandomNumber()
    {
        uniform_int_distribution<int> R();
         return = R;
    }

    float RandomFloat()
    {
        uniform_int_distribution<float> R();
    }

    int RandomNumberRange(int Min =0,int Max =0)
    {
        uniform_int_distribution<int> R(Min,Max);
    }

    float RandomFloatRange(float Min = 0, float Max = 0)
    {
        uniform_int_distribution<float> R(Min,Max);
    }}; 

Answer 1

The random number engine is supposed to be reused, but currently you are declaring it as a local variable, which is never used again and destroyed when the constructor exits. Instead, make the default_random_engine a member of your class.

Then, to get an actual random number from a distribution, you need to call it via its overloaded call operator and pass it the random number engine:

class Random
{
private:
    default_random_engine RandomNumber{time(0)};

public:

    int RandomNumberRange(int Min =0,int Max =0)
    {
        return uniform_int_distribution<int>{Min,Max}(RandomNumber);
    }

    //...
};

There really isn't any reason to return the distribution itself. So I interpreted your question in a way that probably makes more sense.

Also note that the seeding with time(0) is probably a bad one. But this is a huge topic itself. In particular, if you happen to create multiple instances of the Random class in short succession, they will be equally seeded. Instead

default_random_engine RandomNumber{std::random_device{}()};

may work better. It will probably use random numbers provided by the system. (But it may not, for example older versions of MinGW did generate deterministic results this way.)

Answer 2

    int RandomNumber()
    {
        uniform_int_distribution<int> R();
        return = R;
    }

Return type and the type of R in the above function are different. R is of type uniform_int_distribution<int> whereas return type is just int. To return R() you have to change the function return type.

    uniform_int_distribution<int> RandomNumber()
    {
        uniform_int_distribution<int> R;
        return R;
    }

Note that it is not return = R. return is a keyword and not a variable that you can assign to. To return a variable just use return R;.