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Pages: 12
Disch (8348)
+1 @ Helios.

srand is only being passed time()'s return value. Regardless of the optional 'out' parameter time() takes, it still returns the same thing.

Example:

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int main()
{
    time_t t;
    cout << time(0) << endl;
    cout << time(&t) << endl;
    cin.get();
}
 
1351481821
1351481821


As you can see, output is the same with both forms.


To clarify: 't' is not seeding srand(), time()'s return value is. 't' is going unused.
Last edited on
whitenite1 (705)
@Helios
@Disch

Thanks for the time(0) info. I thought I read somewhere that time_t t seeded srand, but I guess I misunderstood. Anyway, using the time(0) will make it easier to remember, when I need to use the srand function.
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