Summary:
The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
bits are filled with zeros.
If E1 has a signed type and non-negative value, and E1 × 2E2 is
representable in the result type, then that is the resulting value;
otherwise, the behavior is undefined.
The code assumed the signed left shift worked like an unsigned left
shift, so make that explicit.
Reviewed By: meyering
Differential Revision:
D3280325
fbshipit-source-id:
178b95ee36b7a1126a79bb825c2ad2ffa6fa9464
inline uint64_t encodeZigZag(int64_t val) {
// Bit-twiddling magic stolen from the Google protocol buffer document;
// val >> 63 is an arithmetic shift because val is signed
- return static_cast<uint64_t>((val << 1) ^ (val >> 63));
+ auto uval = static_cast<uint64_t>(val);
+ return static_cast<uint64_t>((uval << 1) ^ (val >> 63));
}
inline int64_t decodeZigZag(uint64_t val) {