how missing template argument is valid in one but not valid in other function?

why template argument missing in print_all(arr&) is valid but not in f(std::vector&) ?

source code link.

What am I missing ? please explain what happens when we compile ?

template<typename T>
void print_all(T& rhs) 
{
    for(int i = 0; i < rhs.size(); i++) {
        std::cout << rhs[i] << " ";
    }

    std::cout << "\n";
}

void f(std::vector& v) 
{
    std::cout << v[0] << "\n";
}

int main() {
    arr<int, 4> a1 = {4, 5, 6, 7};

    arr<int, 4> a2 = std::move(a1);

    print_all(a2);

    std::vector<int, 4> v = {3, 5, 6, 7};
    f(v);;

}

In this call:

print_all(a2);

The type T of the template parameter rhs is deduced on the call site from the actual argument that you pass (a2). I.e. T is deduced to be arr<int, 4>.
This C++ feature is called Template argument deduction.

On the other hand, the case of your f is different.
f It is not declared to be templated at all but a normal function.
When the compiler compiles it it does not know what kind of std::vector v is supposed to be, and hence the errors.

You can solve it by making f templated:

template<typename T>
void f(std::vector<T>& v) 
{
    std::cout << v[0] << "\n";
}

Live demo

Note:
std::vector does not have a templated parameter for the size, which is dynamic. Therefore in this line:

std::vector<int, 4> v = {3, 5, 6, 7};

v should just be a std::vector<int>:

std::vector<int> v = {3, 5, 6, 7};

(this is already fixed in the live demo link above)