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c++templatesc++11shared-ptr

Deduce shared_ptr T in function declaration

I want to write a function that looks like this:

template<class T>
void Foo(const std::shared_ptr<const T>& ptr);

so that I can call it like this:

std::shared_ptr<int> ptr;
Foo(ptr);

However, the compiler can't deduce T and I have to make it explicit:

Foo<int>(ptr);

or to overload it with void Foo(const std::shared_ptr<T>& ptr).

Can I have a single declaration Foo with const T so that T can be deduced?

Solution

  • Your problem is that ptr is declared as std::shared_ptr<int> while foo requires a std::shared_ptr<const int>. Either you declare ptr with the const qualifier:

    std::shared_ptr<const int> ptr;
Foo(ptr);

    Or you remove the const qualifier from the declaration of foo:

    template<class T>
void Foo(const shared_ptr<T>& ptr);

    Or, unless you really need to know in foo that you are dealing with a shared_ptr, make foo truly generic:

    template<class T>
void Foo(const T& ptr);