> From: openssl-users [mailto:openssl-users-bounces@xxxxxxxxxxx] 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 -- openssl-users mailing list To unsubscribe: https://mta.openssl.org/mailman/listinfo/openssl-users
