Error with return value

Question

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;
}

Eeeeeekkk!!.....Think about it in terms of responsibilities. The method (function) that return random numbers should do only that. The printing should be someone else's responsibility — Mitch Wheat, Dec 07 '13 at 11:59
+1 to counter the unexplained downvote, and because this question is clear and complete, with no reason to downvote — Cheers and hth. - Alf, Dec 07 '13 at 12:26

Domi · Answer 1 · 2013-12-07T12:08:50.817

1

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).

edited Dec 07 '13 at 12:08

answered Dec 07 '13 at 12:02

Domi

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It's highly doubtful that the OP wants the return value from `srand`. – Cheers and hth. - Alf Dec 07 '13 at 12:04
"main" will complain about not returning anything. – Abhineet Dec 07 '13 at 12:04
@Abhineet: no, `main` has a default return value. – Cheers and hth. - Alf Dec 07 '13 at 12:05
I added a return value because I agree with @Abhineet, it's not pretty to not return anything. – Domi Dec 07 '13 at 12:07
@Cheersandhth.-Alf `main` has default return value? I don't think so, but maybe I am wrong, so can you provide a reference link? – hyde Dec 07 '13 at 12:10
@hyde: The C++ standard isn't available for hyperlinking, and it costs quite a bit, but you can get a free draft from the committee pages (or even build it yourself). Then in C++98 through C++11 it's §3.6.1/5. (Sorry, I corrected that which was correct, sort of cross-eyed.) – Cheers and hth. - Alf Dec 07 '13 at 12:18
I googled that up and the first link google threw is :: http://stackoverflow.com/questions/204476/what-should-main-return-in-c-and-c – Abhineet Dec 07 '13 at 12:20

score 1 · Answer 2 · answered Dec 07 '13 at 12:11

#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.

score 0 · Answer 3 · answered Dec 07 '13 at 12:04

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...

score 0 · Answer 4 · answered Dec 07 '13 at 17:15

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.

Error with return value

4 Answers4