> From: openssl-users [mailto:openssl-users-bounces@xxxxxxxxxxx] On Behalf > Of William Roberts > Sent: Wednesday, July 25, 2018 13:00 > > > unsigned char bytes[2]; > > RAND_bytes(bytes, 2); > > return (bytes[0] | (bytes[1] << 8)) & 0x7fff; > > You can ditch the shift logic. Offhand, i'm not sure what would > happen on Big Endian machine, would it leave bit 15 high since it's in > byte 0? No. Bitwise operators in C work according to value, not representation, regardless of the byte order of multibyte integer types in that implementation. > int openssl_rand(void) { > uint16_t x; > RAND_bytes((unsigned char *)&x, sizeof(x)); > return x & 0x7FFF; > } That's OK if you include stdint.h, because you don't care which of the two permissible representations uint15_t has (it has to be pure-binary with no trap representations) - IF your implementation has a 16-bit unsigned integer type. uint16_t won't be defined for an implementation that doesn't. Offhand I don't know of one that is CHAR_BIT 8, though. Personally, I don't care for your version, because I don't like to see code manipulate the representation of an object without specific reason. My version follows the same pattern that correctly-written integer-marshaling code should use, for example; it has the same behavior regardless of implementation details (assuming, once again, that CHAR_BIT is 8). By the way, sizeof is an operator. There's no need to parenthesize its operand, unless the operand is a type. Of course, as Viktor pointed out, this all may be pointless anyway; it's not clear that the OP needs this functionality. -- Michael Wojcik Distinguished Engineer, Micro Focus -- openssl-users mailing list To unsubscribe: https://mta.openssl.org/mailman/listinfo/openssl-users
