Member Avatar for Black Magic

I went to the beach at the weekend and played on one inpeticular game, the one were you pull the lever and XOO or w.e comes out, i was going to make one but thought i would ask if this was the best solution..

#include <conio.h>
#include <iostream>

using namespace std;

int main()
{
    unsigned int firstColum, secondColum, thirdColum;
    
    /////////////////////////////////////////////////////////
    
    firstColum = rand() % 2 + 1;
    
    srand ( time(NULL) );
    
    /////////////////////////////////////////////////////////
    
    secondColum = rand() % 2 + 1;
    
    srand ( time(NULL) );
    
    /////////////////////////////////////////////////////////
    
    thirdColum = rand() % 2 + 1;
    
    srand ( time(NULL) );
    
    /////////////////////////////////////////////////////////
    
    if( firstColum == 1)
    cout << "X";
         else
    cout << "O";
    
    if( secondColum == 1)
    cout << "X";
         else
    cout << "O";
    
    if( thirdColum == 1)
    cout << "X";
         else
    cout << "O";
    
    getch();
}

EDIT: I've noticed i usually get same output

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  • Latest Post Latest Post by Black Magic

Dani AI

Generated

A few focused, practical notes to make the slot-style program reliable and simple to extend.

correctly pointed out why repeated seeding produced identical outputs: reseeding a generator repeatedly (especially with a time-based seed) often resets it to the same state. A more robust approach is to create a single random engine, seed it once, and draw from it for each reel. Prefer the facilities in <random>: they give better statistical properties and let you avoid common pitfalls like modulo bias.

Example pattern (pick an index per reel with one seeded engine):

#include <random>
#include <vector>
#include <string>
#include <iostream>

std::random_device rd;
std::mt19937 eng(rd());
std::vector<std::string> symbols = {"X", "O"};
std::uniform_int_distribution<int> dist(0, symbols.size() - 1);

for (int r = 0; r < 3; ++r)
    std::cout << symbols[dist(eng)];

Why this helps: uniform_int_distribution removes the bias that rand() % n can introduce; std::mt19937 is fast and reproducible if you use a fixed seed (handy for tests); std::random_device will often provide a nondeterministic seed, but its behavior is implementation-dependent — check the docs. See C++ random and uniform_int_distribution for details.

Practical extras: store symbols and payouts in arrays or a map so checking "XXX" or other patterns is just an index lookup; use a loop for any number of reels; use a fixed seed during unit tests to reproduce outcomes; and avoid platform-specific headers if portability matters. These steps make the game easier to maintain and extend for scoring, animation, or additional symbols.

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Member Avatar for Ancient Dragon

>>Is this best solution?
No. srand() should only be called once during the lifetime of the program. Leave lines 12 and 14 but delete lines 18, 20, 24 and 26.

>>#include <conio.h>
That is non-standard function. If you want best solution than delete it.

line 45: delete it. use cin.get() instead.

Member Avatar for Ancient Dragon

lines 20 and 26 are using wrong variable names, so leave them but delete the preceeding calls to srand().

Member Avatar for Black Magic

So are you saying i should only call srand once and use one variable for random number or..?

EDIT: I think i cracked that bit!

#include <iostream>

using namespace std;

int main()
{
    unsigned int firstColum, secondColum, thirdColum;
    
    /////////////////////////////////////////////////////////
    
    srand ( time(NULL) );
    
    firstColum = rand() % 2 + 1;
    
    /////////////////////////////////////////////////////////
    
    if( firstColum == 1)
    cout << "X";
         else
    cout << "O";
    
    firstColum = rand() % 2 + 1;
    
    if( firstColum == 1)
    cout << "X";
         else
    cout << "O";
    
    firstColum = rand() % 2 + 1;
    
    if( firstColum == 1)
    cout << "X";
         else
    cout << "O";
    
    cin.get();
}
Member Avatar for Ancient Dragon

Now, since the code in blocks 17-20 are repeated three times you can use a loop

for(int i = 0; i < 3; i++)
{

firstColum = rand() % 2 + 1;
    
    if( firstColum == 1)
    cout << "X";
         else
    cout << "O";

}
Member Avatar for Black Magic

Sorry for this late reply but i've changed to firstColum, secondColum and thirdColum again, i'll post code when home and i've added options like if output = XXX you win $1, i'm planning to make it $0.10 a go and the total changes etc etc :)

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