4 ms·
If param is unsigned, then "param + 16" cannot overflow; rather, the value wraps around in a language-defined manner. I've been assuming that param is of type i
by _kst_ 5y ago
If param is unsigned, then "param + 16" cannot overflow; rather, the value wraps around in a language-defined manner. I've been assuming that param is of type int (and I stated that assumption).