C++ random numbers logical operator wierd outcome

五迷三道 提交于 2019-12-11 10:16:52

问题


I am trying to make a program generating random numbers until it finds a predefined set of numbers (eg. if I had a set of my 5 favourite numbers, how many times would I need to play for the computer to randomly find the same numbers). I have written a simple program but don't understand the outcome which seems to be slightly unrelated to what I expected, for example the outcome does not necessarily contain all of the predefined numbers sometimes it does (and even that doesn't stop the loop from running). I think that the problem lies in the logical operator '&&' but am not sure. Here is the code:

const int one = 1;
const int two = 2;
const int three = 3;

using namespace std;

int main()
{

    int first, second, third;
    int i = 0;

    time_t seconds;
    time(&seconds);

    srand ((unsigned int) seconds);

    do

    {


    first = rand() % 10 + 1;
    second = rand() % 10 + 1;
    third = rand() % 10 + 1;


    i++;

    cout << first<<","<<second<<","<<third<< endl;
    cout <<i<<endl;
    } while (first != one && second != two && third != three); 

    return 0;
 }

and here is out of the possible outcomes:

3,10,4
1 // itineration variable
7,10,4
2
4,4,6
3
3,5,6
4
7,1,8
5
5,4,2
6
2,5,7
7
2,4,7
8
8,4,9
9
7,4,4
10
8,6,5
11
3,2,7
12

I have also noticed that If I use the || operator instead of && the loop will execute until it finds the exact numbers respecting the order in which the variables were set (here: 1,2,3). This is better however what shall I do make the loop stop even if the order is not the same, only the numbers? Thanks for your answers and help.


回答1:


The issue is here in your condition:

} while (first != one && second != two && third != three);  

You continue while none of them is equal. But once at least one of them is equal, you stop/leave the loop.

To fix this, use logical or (||) rather than a logical and (&&) to link the tests:

} while (first != one || second != two || third != three);  

Now it will continue as long as any of them doesn't match.

Edit - for a more advanced comparison:

I'll be using a simple macro to make it easier to read:

#define isoneof(x,a,b,c) ((x) == (a) || (x) == (b) || (x) == (c))

Note that there are different approaches you could use.

} while(!isoneof(first, one, two, three) || !isoneof(second, one, two, three) || !isoneof(third, one, two, three))



回答2:


You have a mistake in your logical condition: it means "while all numbers are not equal". To break this condition, it is enough for one pair to become equal.

You needed to construct a different condition - either put "not" in front of it

!(first==one && second==two && third==three)

or convert using De Morgan's law:

first!=one || second!=two || third!=three


来源:https://stackoverflow.com/questions/12157137/c-random-numbers-logical-operator-wierd-outcome

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