Deduct template parameter fail while using if constexpr - c++

Question

I am trying to figure out how the sfinae concept works in C++. But I can't convince the object-type compiler if bool is true or false.
#include <iostream>
class A {
public:
void foo() { std::cout << "a\n"; }
};
class B {
public:
void ioo() { std::cout << "b\n"; }
};
template<class T>
void f(const T& ob, bool value = false)
{
if constexpr (!value) ob.foo();
else ob.ioo();
}
int main()
{
A ob1;
B ob2;
f(ob1, true);
f(ob2, false);
}

Answer 1

You need to let the bool parameter to be part of template. In your case, bool value is a run time parameter.
By adding the value as non-template parameter, your code can compile:
template<bool Flag, class T>
// ^^^^^^^^
void f(const T& ob)
{
if constexpr (Flag) {
ob.foo();
}
else
ob.ioo();
}
and you may call like:
f<true>(ob1);
f<false>(ob2);
As a side note, your A::foo() and B::ioo() must be const qualified member functions, as you want to call, with a const objects of each classes inside the f.
That being said, the bool parameter is redundant if you can use the std::is_same
#include <type_traits> // std::is_same
template<class T>
void f(const T& ob)
{
if constexpr (std::is_same_v<T, A>) {
ob.foo();
}
else if constexpr (std::is_same_v<T, B>)
ob.ioo();
}
Now call needs to be just:
f(ob1);
f(ob2);