C++ When template argument deduction fails
Why can't C++ determine that I intend to create a unique_ptr<A>
with this syntax? (a has previ开发者_如何学JAVAously been declared as unique_ptr<A>
)
a = unique_ptr(new A());
It seems awfully redundant to have to include <A>
. This works for most functions templates I use, why not unique_ptr?
EDIT: C++ now supports make_unique, with no redundancy.
std::unique_ptr
is a class template, not a function template. Argument deduction only happens for function templates, not class templates.
A commonly used trick is to write a function template that creates an object of the instantiated class template type, for example:
template <typename T>
std::unique_ptr<T> make_unique_ptr(T* ptr)
{
return std::unique_ptr<T>(ptr);
}
For std::unique_ptr
, though, I'd avoid doing this: a std::unique_ptr
object should directly take ownership of the dynamically allocated object, so there should not be a need for this. Your code should either be written as:
std::unique_ptr<A> a(new A());
or, if a
already exists, a call to reset()
can be used:
a.reset(new A());
As for why type deduction won't work for instantiating a class template, consider the following example:
template <typename T>
struct X
{
template <typename U> X(U) { }
};
There is no way that T
could be deduced from an invocation of the constructor. Even in "simpler" cases where there is a constructor with a parameter of type T
, there can still be trouble since constructors can be overloaded.
精彩评论