How to do this with the rand function?

[szak1592]

You know if I use a rand function in a loop, it may generate the same number twice, how to ensure that it generates distinct numbers? What to add with it.
Here is my code:
void main()
{
int m;
srand(time(NULL));
for (int c=1; c<=5; c++) {
m=rand() % 10 + 1;
cout<<m<<endl;}
getch();
}

[Moschops]

Only call srand(time(NULL)); once in entire program. That's what you are doing, so this is fine.

void main() is just plain wrong, though. C++ insists on int main()

Edit: Do you actually not want the exact same number twice, or do you just want to not repeat the random sequence? In a random sequence, you will get repeats (otherwise it's not random).

[Disch]

2 ways:

1) Create an array that has all the possible outcomes you want (Each number exists only once in the array), then use random_shuffle to shuffle it up, and take the first 5 numbers.

or

2) Keep track of all the numbers you've generated so far (something like a std::set would be a good container to use for this) and keep "rerolling" if you get a number you already got.


#1 is more suitable for small sets, #2 is more suitable for larger sets. Since you are only generating numbers between 1 and 10, I would go with approach #1.




Side notes:

- main should return an int. Always. void main() is incorrect.

- In programming, things are generally zero based, not 1 based. You should try to get used to that.
for (int c=1; c<=5; c++) this loop will work just fine, but is a little strange. A more typical loop would be:
for (int c=0; c<5; c++). It doesn't really matter in this case, but when you start actually using the counter as an index in your loop, starting at 1 will screw you up.

[szak1592]

Okay got it, I will use the array approach. Thanks