Implementing a break in my If-else statements

[PopSmoke]

I'm trying to create a program that rolls a random number 1-6, which is then outputted to a die face, then it stops after each number has been rolled once. I figured majority of the code out, just don't know where to insert my break statements. What should I do?

**CODE BELOW**

// Constants
const int MIN_VALUE = 1; // Min number the die can be
const int MAX_VALUE = 7; // Max number the die can be


// Variables
int randomNumber;

// System time
unsigned seed = time(0);

// Seed the random number generator
srand(seed);
cout << "The number rolled is:" << endl;
randomNumber = (rand() % (MAX_VALUE - MIN_VALUE + 1));
cout << randomNumber << endl << "---------------" << endl;

if (randomNumber == 1) // If a 1 is drawn, output X7
{
OutputData(32);
Sleep(2000);
cout << "---------------" << endl;
}
else if (randomNumber == 2) // If a 2 is drawn, output X1 and X6
{
OutputData(9);
Sleep(2000);
cout << "---------------" << endl;
}
else if (randomNumber == 3) // If a 3 is drawn, output X1, X6 and X7
{
OutputData(41);
Sleep(2000);
cout << "---------------" << endl;
}
else if (randomNumber == 4) // If a 4 is drawn, output X1, X3, X4 and X6
{
OutputData(89);
Sleep(2000);
cout << "---------------" << endl;
}
else if (randomNumber == 5) // If a 5 is drawn, output X1, X3, X4, X6, and X7
{
OutputData(121);
Sleep(2000);
cout << "---------------" << endl;
}
else if (randomNumber == 6) // If a 6 is drawn, output X1, X2, X3, X4, X5, and X6
{
OutputData(95);
Sleep(2000);
cout << "---------------" << endl;
}
}

[Duthomhas]

I am unsure what you are asking. Do you want to roll the die more than once?

[PopSmoke]

No. I want the loop to stop once 1,2,3,4,5, and 6 were each rolled once, respectively.

[Duthomhas]

Duthomhas wrote:

Do you want to roll the die more than once?

PopSmoke wrote:

No.

PopSmoke wrote:

I want the loop to stop once 1,2,3,4,5, and 6 were each rolled once, respectively.

That’s more than once.

[PopSmoke]

my bad, I thought you meant for each number

[Duthomhas]

So, what exactly do you want your program to do? Produce the numbers 1–6 in random order?

[PopSmoke]

Yeah, then stop once each number has been called once in the loop.

-stop the program automatically after each of the outputs is displayed at least one time (1-6)

[lastchance]

If you want EXACTLY once, then just put the numbers in an array and use std::shuffle()

If you want UNTIL AT LEAST ONCE then, e.g.,

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <vector>
using namespace std;

int main()
{
   srand( time( 0 ) );
   const int LOW = 1, HIGH = 6;
   vector<bool> available( 1 + HIGH, true );
   int remaining = HIGH - LOW + 1;
   
   while( remaining )
   {
      int x = LOW + rand() % ( HIGH - LOW + 1 );
      remaining -= available[x];
      available[x] = false;
      cout << x << '\n';;
   }
}

2
4
3
2
5
2
6
4
2
1

[seeplus]

Do you mean something like this - each number in the designated range will only be visited once in a random order:

#include <iostream>
#include <random>
#include <algorithm>
#include <numeric>

void OutputData(unsigned n) { std::cout << " output " << n << '\n'; }	// For testing

std::mt19937 rng(std::random_device {}());

int main() {
	constexpr unsigned MIN_VALUE {1}; // Min number the die can be
	constexpr unsigned MAX_VALUE {6}; // Max number the die can be
	constexpr unsigned odata[MAX_VALUE - MIN_VALUE + 1] {32, 9, 41, 89, 121, 95};

	unsigned nos[MAX_VALUE - MIN_VALUE + 1];

	std::iota(std::begin(nos), std::end(nos), 1);
	std::shuffle(std::begin(nos), std::end(nos), rng);

	for (auto n : nos) {
		std::cout << "The number rolled is: " << n;
		OutputData(odata[n - MIN_VALUE]);
	}
}

The number rolled is: 1 output 32
The number rolled is: 6 output 95
The number rolled is: 5 output 121
The number rolled is: 2 output 9
The number rolled is: 3 output 41
The number rolled is: 4 output 89

[lastchance]

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <vector>
using namespace std;

void output( int p )
{
   char bits[] = { 0, 7, 6, 1, 5, 2, 4, 8, 3 };
   cout << "\n\n";
   for ( int i = 0; i < 9; i++ ) cout << ( p & (1<<bits[i]) ? '*' : ' ' ) << ( ( i + 1 ) % 3 ? "" : "\n" );
   cout << "\n\n";
}


int main()
{
   int dice[] = { 0, 32, 9, 41, 89, 121, 95 };
   srand( time( 0 ) );
   const int LOW = 1, HIGH = 6;
   vector<bool> available( 1 + HIGH, true );
   int remaining = HIGH - LOW + 1;
   
   while( remaining )
   {
      int x = LOW + rand() % ( HIGH - LOW + 1 );
      remaining -= available[x];
      available[x] = false;
      // cout << x << '\n';
      output( dice[x] );
   }
}

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[jonnin]

that is a slick way to do the output, but standard 3 is the diagonal.
the rest match a standard d6.

[lastchance]

Thanks @jonnin. I have edited to accommodate that. But there are still various possible orders (because the diagonal could go the other way, for example).

I just wondered why the student was outputting such odd numbers.

[seeplus]

As the output relates to X1 - X7, possibly it is to interface to some hardware? 7-seg display?

[seeplus]

if the requirement is to continue until at least one of each number is determined, then consider:

#include <iostream>
#include <random>
#include <bitset>

void OutputData(unsigned n) { std::cout << " output " << n << '\n'; }	// For testing

std::mt19937 rng(std::random_device {}());

int main() {
	constexpr unsigned MIN_VALUE {1}; // Min number the die can be
	constexpr unsigned MAX_VALUE {6}; // Max number the die can be
	constexpr unsigned odata[MAX_VALUE - MIN_VALUE + 1] {32, 9, 41, 89, 121, 95};
	const std::uniform_int_distribution<unsigned> distrib(MIN_VALUE, MAX_VALUE);
	std::bitset<MAX_VALUE - MIN_VALUE + 1> bits;

	while (!bits.all()) {
		const auto val {distrib(rng)};

		bits.set(val - MIN_VALUE);
		std::cout << "The number rolled is: " << val;
		OutputData(odata[val - MIN_VALUE]);
	}
}

The number rolled is: 3 output 41
The number rolled is: 3 output 41
The number rolled is: 2 output 9
The number rolled is: 2 output 9
The number rolled is: 1 output 32
The number rolled is: 4 output 89
The number rolled is: 5 output 121
The number rolled is: 1 output 32
The number rolled is: 5 output 121
The number rolled is: 3 output 41
The number rolled is: 6 output 95

[PopSmoke]

@seeplus you're correct. Those numbers I used correlate to a TDB module which holds a die face and a 7 segment display