2

srand(0) and srand(1) give the same results. srand(2), srand(3), etc. give different results.

Any reason why seed = 0 and seed = 1 yield the same random sequence?

Can't find an explanation in the man page. Only that if a seed is not provided, seed = 1 is used.

Thanks.

João M. S. Silva
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    Apparently 0 means no seed and, as you state, leads to seed == 1. – mah Dec 09 '14 at 18:48
  • why they should be different? – yeyo Dec 09 '14 at 18:48
  • @mah: I think no seed means `srand()` was not called. – João M. S. Silva Dec 09 '14 at 19:44
  • OP: please try searching first. If you'd typed `srand` into the search box, the first result would have answered your question. – DoxyLover Dec 09 '14 at 19:53
  • @DoxyLover: I searched before and during the posting (among the suggestions) but with a search string that did not return the question you mention and which this one is a duplicate of. I also searched with Google. What seems obvious is not always right. – João M. S. Silva Dec 09 '14 at 21:06
  • @JoãoM.S.Silva since the statement came from the srand() man page, that would seem a bit silly (but I'll agree that the statement itself is somewhat strange since the function has a required argument). Beyond that, considering the source review provided by slugonamission I think you should perhaps reconsider your thought though. – mah Dec 10 '14 at 12:43
  • @mah: I can't fully understand what you say, but the man page for srand is the same for rand and rand_r, and what it says is that "If no seed value is provided, the rand() function is automatically seeded with a value of 1". So I was led to think that `srand(1)` was the default but `srand(x)` with `x` other than 1 would lead to a different result. – João M. S. Silva Dec 10 '14 at 15:40

4 Answers4

11

Within the glibc sources for srandom_r (which is aliased to srand), line 179: 

/* We must make sure the seed is not 0.  Take arbitrarily 1 in this case.  */
if (seed == 0)
    seed = 1;

It's just an arbitrary decision basically.

slugonamission
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1

Depends on compiler!

srand(0);
int a=rand(),b=rand();
srand(1);
int c=rand(),d=rand();

VC 2005 result:

a    0x00000026    int
b    0x00001e27    int
c    0x00000029    int
d    0x00004823    int
Anonymous
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1

This is an implementation dependent behaviour.

For instance, POSIX.1-2001 gives the following example of an implementation of rand() and srand()

static unsigned long next = 1;

/* RAND_MAX assumed to be 32767 */
int myrand(void) {
    next = next * 1103515245 + 12345;
    return((unsigned)(next/65536) % 32768);
}

void mysrand(unsigned seed) {
    next = seed;
}

Now, if you use this implementation you will end up with:

0
16838

for srand(0) and srand(1) respectively.

ref.: http://linux.die.net/man/3/rand

I ran into a quite similar problem before, where rand() yielded different sequences for the same seed across different platforms. Lesson learned, portable code should implement his own PRNG.

yeyo
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-2

The function srand() is used to initialize the pseudo-random number generator by passing the argument seed.

So if the seed is set to 1 then the generator is reinitialized to its initial value. Then it will produce the results as before any call to rand and srand.

so srand(1) actually represent the result srand(0).

Hex.Shadow
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    Fantastic, but how does this in any ay address the OP's question of **why** both 0 and 1 produce the **same** prng sequence? – WhozCraig Dec 09 '14 at 18:52