srand(0) and srand(1) give the same results?

Question

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.

Answer 1

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.

Answer 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).

Answer 3

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

Answer 4

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.