CODEWITHSUNDEEP
c++
user3054263
user3054263

Reputation: 39

Error with return value

So I am creating a random number generator and I keep getting a problem. In my program I have it print out the code into the "int main()" function. The problem is that it prints a 0 after. It's also saying I have to use return, but I don't want to. I want to keep the random number generator in another function because I will be adding much more in the future.

#include<iostream>
#include<cstdlib>
#include<ctime>

using namespace std;

int randRoll();

//Calls the other function
int main()
{
    cout << randRoll() << endl;

    system("Pause");
}

//Gets the random number
int randRoll()
{
    srand(time(0));

    for (int x = 1; x <= 1; x ++)
    {
        cout <<"Your random number is " << 1 +(rand()%4) << endl;
    }

    return 0;
}

Upvotes: 2

Views: 82

Answers (4)

Matthieu M.
Matthieu M.

Reputation: 300419

Well, of course it prints two numbers, you asked it to!

// This prints "Your random number is [1-4]\n"
cout <<"Your random number is " << 1 +(rand()%4) << endl;

// This prints "0\n" since "randRoll()" returns 0
cout << randRoll() << endl;

Your main function could just be int main() { randRoll(); } because randRoll already prints something.

Upvotes: 0

Cheers and hth. - Alf
Cheers and hth. - Alf

Reputation: 145457

#include<iostream>
#include<stdlib.h>
#include<time.h>

using namespace std;

//Gets the random number
int randRoll()
{
    return 1 +(rand()%4);
}

//Calls the other function
int main()
{
    srand( static_cast<unsigned>( time(0) ) );

    for (int x = 1; x <= 7; x ++)
    {
        cout <<"Your random number is " << randRoll() << endl;
    }
}

Salient points:

  • Preferably use e.g. <stdlib.h>, not <cstdlib>, to avoid some problems.
  • Preferably avoid forward declarations of functions (also avoids some problems).
  • Call srand only once.

Upvotes: 1

Domi
Domi

Reputation: 24648

Try this instead:

#include<iostream>
#include<cstdlib>
#include<ctime>

using namespace std;

int randRoll();

// entry point
int main()
{
    srand(time(0));     // initialize randomizer

    cout << randRoll() << endl;

    system("Pause");

    return 0;
}

//Gets the random number
int randRoll()
{
    auto x = 1 +(rand()%4);

    // do something with x here (but don't call cout!)

    return static_cast<int>(x);
}

The problem you have is that you are not returning the randomly generated value (which I called x in my code). Also, you were trying to print out the randomly generated value twice (but incorrectly).

Upvotes: 1

hyde
hyde

Reputation: 62906

cout << randRoll() << endl;

That prints out return value of that function, which is 0. How could not, when you tell it to print it? Well, like this, by not asking it to print return value:

randRoll(); // prints some random numbers
cout << endl; // prints newline

Then return value is ignored, and you can change it to void type if you want.

Note that as a whole, doing it like this might not make sense, but I think that answers your question and allows you to move on to next problem...

Upvotes: 0

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