I am trying to make template wrapper function, that should forward parameters and return value. And I can't decide what is better to use auto&&
or decltype(auto)
for return type. I've read Scott Meyers article and understood that it is necessary to return decltype(auto)
compared to auto
not to strip ref_qualifiers.
As far as I understand the same argument works for using auto&&
over auto
.
Now I have following questions:
decltype(auto)
and auto&&
when we return reference to object?rvalue
object, like: return int{};
? Will return value be dangling reference?decltype(auto)
and auto&&
? What better fits as forward return type?'auto' lets you declare a variable with a particular type whereas decltype lets you extract the type from the variable so decltype is sort of an operator that evaluates the type of passed expression.
In C++14, you can just use auto as a return type.
What is the difference between
decltype(auto)
andauto&&
?
decltype(auto)
covers three cases. When returning lvalues, the return type would be T&
(lvalue-reference); for xvalues, the return type would be T&&
(rvalue-reference); for prvalues, the return type would be T
(non-reference, i.e. return by-value).
auto&&
covers only two cases. When returning lvalues, the return type would be T&
(lvalue-reference); for rvalues, including xvalues and prvalues, the return type would be T&&
(rvalue-reference). (Forwarding reference is always a reference.)
What happens if we return rvalue object, like:
return int{};
? Will return value be dangling reference?
For auto&&
the return type is rvalue-reference, so yes, the returned reference is always dangling. For decltype(auto)
the return type is non-reference then no such trouble.
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