What‘s the difference between srand(1) and srand(0

Question

I just found out the hard way that srand(1) resets the PRNG of C(++) to the state before any call to srand (as defined in the reference). However, the seed 0 seems to do the same, or the state before any call to srand seems to use the seed 0. What’s the difference between those two calls or what is the reason they do the same thing?

For example this code (execute on Ideone)

#include <stdio.h>
#include <stdlib.h>

int main() {
    for (int seed = 0; seed < 4; seed++ ) {
        printf( "Seed %d:", seed);
        srand( seed );
        for(int i = 0; i < 5; i++ )
            printf( "    %10d", rand() );
        printf( "\n");
    }
    return 0;
}

returns

Seed 0:    1804289383     846930886    1681692777    1714636915    1957747793
Seed 1:    1804289383     846930886    1681692777    1714636915    1957747793
Seed 2:    1505335290    1738766719     190686788     260874575     747983061
Seed 3:    1205554746     483147985     844158168     953350440     612121425

Answer 1

It is probably an implementation detail. The standard mandates that the random seed 1 is special, and the internal register of your specific random generator algorithm is probably zero-initialized, thus causing the same random sequence for seed(0) and seed(1). I'd even wager that the first line of your srand() implementation looks like:

if ( seed == 1 ) seed = 0;

to force standard-conformant behaviour.

Generally, the random number generators for rand() and srand() are not required to give different sequences for different seeds, but the same sequence for the same seed. So, don't rely on different seeds generating different random sequences, and you should be fine. If not, welcome to implementation-specific fun.