12 ms·
> that helpful "&arg" goes away and it's just "arg". You still get a reminder it's a pointer by the * and -> notation.
by kevin_thibedeau 4y ago
> that helpful "&arg" goes away and it's just "arg".
You still get a reminder it's a pointer by the * and -> notation.
- cpgxiii 4y agoThat only goes so far - without reasonable method names you still don't know if the operation actually mutates the parameter, e.g. in void func(ParamType* param) { ... auto res_1 = param->Operation1(1); auto res_2 = param->Operation2(2); ... } the presence of -> doesn't tell you which method mutates. The only thing that would tell you that would be knowing which method is const.