3 ms·
Sure, but in the scenario you describe, (count-1)*(count-2) also overflows, and that's why the generated code already to 64 bits for the multiply. (See my cous
by valleyer 2y ago
Sure, but in the scenario you describe, (count-1)*(count-2) also overflows, and that's why the generated code already to 64 bits for the multiply. (See my cousin comment to this one.)