Call one class function in another class

Question

I'm having troubles with classes. I have a class that generates random numbers, and I want to use this random number generator in another class to create a Powerball simulator.

This is my .cpp file for the random number generator:

#include "RandomNumber.h" #include <random> #include <utility> using namespace std; RandomNumber::RandomNumber( int min, int max, bool minInclusive, bool maxInclusive ) : mMinimum( min ), mMaximum( max ) { if (mMinimum > mMaximum) { swap( mMinimum, mMaximum ); } if (!minInclusive) { mMinimum++; } if (!maxInclusive) { mMaximum--; } } int RandomNumber::random( ) { static random_device rd; static mt19937 generator(rd()); uniform_int_distribution<> distro( mMinimum, mMaximum ); return( distro( generator ) ); }

This is my header file for the random number generator:

#ifndef RandomNumber_h #define RandomNumber_h #include <stdio.h> class RandomNumber { public: RandomNumber( int min, int max, bool minInclusive = true, bool maxInclusive= true ); // supply a number between min and max inclusive int random( ); private: int mMinimum, mMaximum; }; #endif /* RandomNumbers_h */

I wanted to call the member function in another class called PowerballLottery and store the 6 values if they're within the appropriate range, I tried to use

RandomNumber.random( 1, 69 )

and

RandomNumber::random( 1, 69 )

but neither worked. I'm wondering what is the correct syntax. Thank you so much for reading this post.

Michael Albers · Accepted Answer · 2016-02-23 04:37:49Z

2

To access a member function you need to instantiate the class.

RandomNumber randNum(1, 69); int newRandomNumber = randNum.random();

answered Feb 23, 2016 at 4:37

Michael Albers

3,7993 gold badges23 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Aposhian Over a year ago

True, but in this context, it would make more sense for random(...) to just be static member function of PowerballTicket. If you need to create dummy instances just to use a method, it is a sign that some static action is in order...

aRandomStudent Over a year ago

Thank you so much it worked! If I want to generate a random number within another range in the same function, do I just call RandomNumber randNum (1, 23); like this? Or do I need a new instance?

Michael Albers Over a year ago

@Yvonne Yes, you'll need a new instance.

James Adkison · Accepted Answer · 2016-02-23 04:38:43Z

The function signature of the random function is:

int RandomNumber::random();

This returns an int and does not take any parameters. Therefore calling random(1, 69) is wrong.

Furthermore, the random function isn't static. Therefore calling RandomNumber::random(1, 69) is wrong (also because it doesn't accept any parameters).

In order to call the random function you need to instantiate an object and then call the function using the object instance. For example:

RandomNumber prng(1, 69); prng.random();

Weak to Enuma Elish · Accepted Answer · 2016-02-23 04:38:19Z

RandomNumber is the name of a class, not an instance, and random is a non-static member function. To call random as it currently is, you would have to create an instance of RandomNumber.

RandomNumber rn(/*...args...*/); rn.random(); //or RandomNumber(/*...args...*/).random();

You could instead make random a static member function (or just not a member at all) that takes the constructor parameters as arguments instead.

Ethan Fischer · Accepted Answer · 2016-02-23 04:43:40Z

Also, don't forget to add

#include "RandomNumber.h"

to the top of your PowerBall.cpp. Otherwise PowerBall won't know what RandomNumber is!

Aposhian · Accepted Answer · 2016-02-23 04:52:25Z

Your problem is that since random() is a member function of RandomNumber and so it can only be called when you instantiate an instance of that class, using the constructor to make an actual object. If you don't really need a specific instance of that class (which it looks like you don't) you can make the function static:

static int random();

This way it can be called from other classes or main code (as long as Random.h is included) just by using RandomNumber::random(), because static member functions don't depend on specific instances of a class. See http://en.cppreference.com/w/cpp/language/static

Essentially, a static doesn't have the implicit parameter this, a pointer to the instance of the object that a regular member function is called on:

x.useSomeMethod(); // then inside the call of useSomeMethod(), *this == x

However, for your application, the RandomNumber class doesn't seem like it needs to be its own class at all. What you can do if you want to use it in PowerballTicket, but not make it available to the code that uses that class, is to make random() a private member function of PowerballTicket, and this can also be static since (I'm assuming) it doesn't care what instance of PowerballTicket it is being used inside:

class PowerballTicket{ public: PowerballTicket(int x1, int x2, int x3, int x4, int x5, int x6); int getBall1(); int getBall2(); int getBall3(); int getBall4(); int getBall5(); int getPowerball(); private: static int random(int min, int max, bool minInclusive = true, bool maxInclusive = true); int mBall1; int mBall2; int mBall3; int mBall4; int mBall5; int mPowerball; };

But I'm not sure where you are actually planning on calling random() from the code you gave, so you may want to modify this as needed.

Collectives™ on Stack Overflow

Call one class function in another class

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Related