为什么编译器不能从返回类型中推导出模板参数？

Question

Given the following code

#include <vector>
#include <memory>

using namespace std;

class MyBase
{};

class MyDerived : public MyBase
{};

template<class Base, class Derived>
vector<Base> makeBaseVec(const Derived& obj, const typename vector<Base>::size_type size)
{
    vector<Base> out;
    for (typename vector<Base>::size_type i = 0; i < size; i++)
    {
        out.push_back(Base(obj) /* copy constructor */);
    }

    return out;
}

int main()
{
    MyDerived a;
    vector<MyBase> v = makeBaseVec<MyBase>(a, 10);
}

Live example

Why do I get the error

main.cpp:13:14: note:   template argument deduction/substitution failed:
main.cpp:29:41: note:   couldn't deduce template parameter 'Base'
     vector<MyBase> v = makeBaseVec(a, 10);
                                         ^

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

I can rectify this by changing line 27 to

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

but this felt unnecessary.

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.