I have a for
loop that runs 15 times, with dh.setDoors()
in every iteration.
What setDoors
does is call srand(time(0))
, then whenever a random number is needed it'll use, for example, carSetter = rand()%3+1
. Alternatively, it may use decider = rand()%2+1
.
Now, normally decider
and carSetter
are used in a different ways, but I suspected a problem and made it print out carSetter
and decider
at every iteration. Here's what came out:
Door 1 has car
Decider is 2
Door 1 has car
Decider is 2
Door 1 has car
Decider is 2
Door 1 has car
Decider is 2
Door 1 has car
Decider is 2
etc...
The values '1' and '2' change when I run it multiple times, but are still the same throughout the 15 times.
Since the loop is running 15 different times, shouldn't carSetter
and decider
print out a different random number every iteration?
When I don't have srand(time(0))
, it works as expected, but there's no seed set, so it's the same sequence of "random" numbers each time, so it's probably a problem with the seed?
random seed() example to generate the same random number every time. If you want to generate the same number every time, you need to pass the same seed value before calling any other random module function.
The RAND function in stand-alone applications generates the same numbers each time you run your application because the uniform random number generator that RAND uses is initialized to same state when the application is loaded.
rand() returns a pseudo-random number in the range of [0, RAND_MAX). RAND_MAX: is a constant whose default value may vary between implementations but it is granted to be at least 32767.
If you use randomNumber() multiple times in your program it will generate new random numbers every time. You can think of each randomNumber() like a new roll of a die.
When you call srand(x)
, then the value of x
determines the sequence of pseudo-random numbers returned in following calls to rand()
, depending entirely on the value of x
.
When you're in a loop and call srand()
at the top:
while (...) {
srand(time(0));
x = rand();
y = rand();
}
then the same random number sequence is generated depending on the value that time(0)
returns. Since computers are fast and your loop probably runs in less than a second, time(0)
returns the same value each time through the loop. So x
and y
will be the same each iteration.
Instead, you only usually need to call srand()
once at the start of your program:
srand(time(0));
while (...) {
x = rand();
y = rand();
}
In the above case, x
and y
will have different values each time through the loop.
Every time you invoke srand(time(0))
, you're seeding the pseudo-random number generator, imbuing it with a new pseudo-random sequence of numbers. The sequence is different depending on what the argument to srand
is, and in this instance you use time(0)
so, assuming you call your program at most once per second, you'll always get a new sequence. When you call rand()
, you just get the next number in this sequence.
However, since you've decided to call srand
multiple times in your program, and because your program is fast (i.e. time(0)
isn't changing), all you're doing is repeatedly resetting the PRNG to the same sequence. That's why you're always getting the same values - you keep reseeding the PRNG to be the same sequence, and this also moves the cursor to the beginning of the sequence.
Seed once. Once.
Like Kerrek said, seeding is done only once, in the beginning of the program. After one srand call rand as many times as you like.
You need to run srand
in the beginning of your program, for example in the main
-function.
When you run srand(time(0))
at the top of the function in which you are using rand()
, you are likely to give it the same seed every time. time(0)
gives the time in seconds, so you would need to avoid calling setDoors
twice within the same second in order to get different numbers every time. If you do call setDoors
twice within the same second, the random seed will be the same, and subsequent calls to rand()
will generate the same sequence of pseudorandom numbers.
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