4 ms·
> `a[i++] = a[i++] + 1;` This is probably not what you want, as "i" is increased twice.
by vardump 3y ago
> `a[i++] = a[i++] + 1;`
This is probably not what you want, as "i" is increased twice.
- jrumbut 3y agoThat's why it matters in that situation, because it changes what actually happens (not just the performance).