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

How glibc does it:

around line 181 of glibc/stdlib/random_r.c, inside function __srandom_r

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

But that's just how glibc does it. It depends on the implementation of the C standard library.