What happens when you use srand with no arguments?

Question

The question is pretty straightforward. What exactly happens when you use "srand" with no arguments? What is the expected behavior?

srand();
my $x1 = int(rand(65536)) % 65536;
my $x2 = int(rand(65536)) % 65536;
print "$x1\n";
print "$x2\n";

`man srand` sayeth `If no seed value is provided, the rand() function is automatically seeded with a value of 1.` (presumably applies to srand too) — Barry Carter, Aug 14 '22 at 13:13
Possible duplicate of https://stackoverflow.com/questions/21273550/how-does-srand-relate-to-rand-function — JRFerguson, Aug 14 '22 at 13:37
What do you think is happening? Note that you already have a number below 65536, so you don't need to take the modulus. — brian d foy, Aug 14 '22 at 15:48

score 2 · Answer 1 · answered Aug 14 '22 at 13:51

2

According to the documentation:

When called with a parameter, "srand" uses that for the seed; otherwise it (semi-)randomly chooses a seed.

answered Aug 14 '22 at 13:51

jhnc

1

You're right and I'm wrong. I missed the `perl` tag – Barry Carter Aug 14 '22 at 13:52

1 Answers1