I think the question is clear enough. Will the auto
keyword auto-detect const-ness, or always return a non-const type, even if there are eg. two versions of a function (one that returns const
and the other that doesn't).
Just for the record, I do use const auto end = some_container.end()
before my for-loops, but I don't know if this is necessary or even different from normal auto
.
const auto x = expr;
differs from
auto x = expr;
as
const X x = expr;
differs from
X x = expr;
So use const auto
and const auto&
a lot, just like you would if you didn't have auto
.
Overload resolution is not affected by return type: const
or no const
on the lvalue x
does not affect what functions are called in expr
.
Maybe you are confusing const_iterator
and const iterator
. The first one iterates over const elements, the second one cannot iterate at all because you cannot use operators
++ and -- on it.
Note that you very seldom iterate from the container.end()
. Usually you will use:
const auto end = container.end();
for (auto i = container.begin(); i != end; ++i) { ... }
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