undefined | Better HN

0 pointskevincox4y ago0 comments

How does this work in C++ with operator overloading. Are they still the same? That would make for some interesting obfuscated code.

0 comments

No, they're not. It works by: if either a or b in a[b] is a class/enumeration type, call a.operator[](b).

> How does this work in C++ with operator overloading.

you can't overload int::operator[](...)

I guess the question was whether the subscript operator is always assumed commutative in C++.

j / k navigate · click thread line to collapse

0 pointskevincox4y ago0 comments

How does this work in C++ with operator overloading. Are they still the same? That would make for some interesting obfuscated code.

No, they're not. It works by: if either a or b in a[b] is a class/enumeration type, call a.operator[](b).

> How does this work in C++ with operator overloading.

you can't overload int::operator[](...)

I guess the question was whether the subscript operator is always assumed commutative in C++.

j / k navigate · click thread line to collapse