> This is called a mapping from int to int that means, if two functions stamped from the same template have the same mapping, then the same input gives the same output.
> But with current C++ it is not the case.
> Example:
> template<int I>
> int foo (int j)
> {
> return I * j;
> }
>
> int main()
> {
> auto f1 = &foo<5>;
> auto f2 = &foo<-3>;
>
> if(std::is_same_v<decltype(f1), decltype(f2)>)
> cout << " f1(1) must equal f2(1) is" << boolalpha<< (f1(1)==f2(1));
> }
>
> Try this snippet, and reason about it. That's all what i can say.
But `foo<5>` and `foo<-3>` do not "have the same mapping" of `int` to
`int`. As such, there is no reason to expect `f1(1)` and `f2(1)` to be
equal.
Two instantiations of the same template with different parameters
generate different things. That's *how templates work*.`foo<5>` and
`foo<-3>` are expected to be different functions with different
behavior. Two different functions that have the same type are still
*two different functions*.
----Hello Jason,
I guess you get exactly what i mean, you just want to argue.
You say f1 has the same type as f2 , right?
But at the same time they give me two different values.
It is like you tell me
int(1) != int(1) // same type different values.
But I guess i understand what you told me once, a while back ago.
"Put 20 C++ expert in a room to fix a problem you will get 20 opinions and no solution"