On Friday, 30 September 2016 at 05:01:52 UTC, Steven
Schveighoffer wrote:
On 9/29/16 9:48 PM, Jacob wrote:
It still requires the &, what it prevents is this situation:
auto pValue = expr; // wanted pointer, expr evaluates to
non-ptr value
through change of code or simply forgetting "&"
Wait, what happens when you do that? If that's now an error,
this is a non-starter.
-Steve
Jacob is simply asking that you may explicit the fact that you
expect the right hand side of the assignment to evaluate to a
pointer while still using the type deduction offered by 'auto'.
auto p = expr;
Depending on the return type of expr, p's type can be either a
pointer or a copy.
auto *p = expr;
Imposes that expr evaluate to a pointer.
Although the former is valid in C++ whatever expr evaluates to, I
always use the latter when I expect expr to evaluate to a
pointer. It also add consistency with 'auto &'.
