Why can't the compiler deduce template parameter from return type?

Question

Given the following code

#include 
#include 

using namespace std;

class MyBase
{};

class MyDerived : public MyBase
{};

template&l         


        

Answer 1

Shouldn't the compiler be able to deduce the the template parameter Base from the type of v ?

The type of v is not considered by the template type deduction mechanism when you call makeBaseVec. What if you were to call the function and discard the return value?

Return types do not participate in type deduction or overload resolution.

If you don't want to repeat yourself, you can use type deduction on v instead:

auto v = makeBaseVec<MyBase>(a, 10);

In fact, almost always auto is a good policy for variables.