3 ms·

For 32-bit types, the undefined behavior is where I expected it to be, but it is the opposite for the 16-bit types. Why are the 16-bit types different? To make

by ryao 3y ago

For 32-bit types, the undefined behavior is where I expected it to be, but it is the opposite for the 16-bit types. Why are the 16-bit types different? To make things weirder, I tried 8-bit and 64-bit versions. The 64-bit version behaved like the 32-bit versions, but the 8-bit versions had no undefined behavior checks. Why is 8-bit so different?

dzaima 3y ago
Smaller-than-'int' types get implicitly promoted to 'int' before arithmetic. Results are indeed very funky.
- ryao 3y ago
  That makes sense then. Multiplying two uint16_t can wraparound in an int, while two int16_t or smaller cannot.