> From: openssl-users [mailto:[email protected]] On Behalf Of
> Sudarshan Soma
> Sent: Wednesday, July 25, 2018 12:13
> But rand() returns max value of 32767 . Is there a recomended way to
> convert RAND_bytes to libc rand()
> something like this?
> unsigned char buf[2];
> RAND_bytes(buf,2)
> int *rndp = malloc(4);
> memcpy(rndp,buf,2);
> return (unsigned) ((*rndp) % 32768)
Ugh. Memory leak, unnecessary malloc, undefined behavior (only part of the rdnp
object is initialized)... I really hope you don't have code like this in your
application.
C guarantees unsigned integer types use a pure binary representation, and 32767
is 2**15 - 1. So assuming you're only using octet-based C implementations
(limits.h defines CHAR_BIT as 8), which is very likely the case, just do this:
unsigned int openssl_rand(void) {
unsigned char bytes[2];
RAND_bytes(bytes, 2);
return (bytes[0] | (bytes[1] << 8)) & 0x7fff;
}
Untested, but I think that will work on any conforming C implementation with
CHAR_BIT == 8, and as long as the 15 least-significant bits of the output of
RAND_bytes are unbiased, the result will be an unbiased value in [0,32767].
Note this does not give you the semantics of C's rand, as it ignores any
invocation of srand. Some C programs require a predictable rand; they use it
for reproducible Monte Carlo test runs, for example. So replacing rand this way
is not necessarily valid.
Also, calling it "rand" would be a violation of the C specification, so if you
want your C applications to conform to the spec, you'll have to change them
anyway. Or use a macro, provided the application code never suppresses a macro
definition for rand.
--
Michael Wojcik
Distinguished Engineer, Micro Focus
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